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# RBSE Solutions for Class 12 Physics Chapter 8 Magnetism and Properties of Magnetic Substances

## Rajasthan Board RBSE Class 12 Physics Chapter 8 Magnetism and Properties of Magnetic Substances

RBSE Class 12 Physics Chapter 8 Text Book Exercise with Answers

### RBSE Class 12 Physics Chapter 8 Multiple Choice Type Questions

Question 1.
If the distance between two magnetic poles of unit magnetic strength is 1 m, then force acting between them will be:
(a) 4π × 10-7 N
(b) 4π N
(c) 10-7 N
(d) $$\frac{4 \pi}{10^{-7}}$$ N
Answer:
(c) 10-7 N

Question 2.
Magnetic susceptibility for superconductors is :
(a) +1
(b) -1
(c) zero
(d) infinity
Answer:
(b) -1
For superconductors, value of magnetic susceptibility is -1.

Question 3.
Magnetic susceptibility of free space is :
(a) +1
(b) -1
(c) zero
(d) infinity
Answer:
(c) zero
Magnetic susceptibility of free space is zero.

Question 4.
Magnetic susceptibility is negative and very less for :
(a) ferromagnetic substances
(b) paramagnetic substances
(c) diamagnetic substances
(d) all of the above
Answer:
(c) diamagnetic substances
For diamagnetic substances.

Question 5.
If the relative permeability of a substance is 1.00001, then substance will be :
(a) ferromagnetic
(b) paramagnetic
(c) diamagnetic
(d) none of these
Answer:
(b) paramagnetic
As μr > 1, so substance should be paramagnetic.

Question 6.
Unit of magnetic moment is :
(a) Wb
(b) Wb/m2
(c) A/m
(d) Am2
Answer:
(d) Am2
Unit of magnetic moment is Am2

Question 7.
Wb × A/m is equals to :
(a) J
(b) N
(c) H
(d) W
Answer:
(b) N
Wb × $$\frac{\mathrm{A}}{\mathrm{m}}$$ is equal to N.

Question 8.
In which of the following, the magnetic field does not interplay with :
(a) magnet
(b) accelerated magnet
(c) static charge
(d) dynamic charge
Answer:
(c) static charge
Magnetic field does not interact with static charges.

Question 9.
The cause of dimegnetism is :
(a) orbital motion of electrons
(b) spin motion of electrons
(c) paired electrons
(d) none of the above
Answer:
(a) orbital motion of electrons
Due to orbital motion of electrons.

Question 10.
Magnetic moment of diamagnetic substances is :
(a) infinity
(b) zero
(c) 100 Am2
(d) none of these
Answer:
(b) zero
Magnetic moment of diamagnetic substance is zero.

Question 11.
The relative permeability of ferromagnetic substance is :
(a) μr > 1
(b) μr >> l
(c) μr = 1
(d) μr = 0
Answer:
(b) μr >> l
For ferromagnetic substance μr >> l .

Question 12.
Vertical component of earth’s magnetic field is zero at:
(a) magnetic pole
(b) geographical pole
(c) magnetic meridian
(d) none of these
Answer:
(d) none of these
None of the option is correct.

Question 13.
Area of hysteresis loop of any substance represent:
(a) energy loss in magnetising a substance in unit cycle
(b) energy loss in magnetising unit volume in unit cycle
(c) energy loss in magnetising a substance for unit volume
(d) energy loss in magnetising a substance
Answer:
(d) energy loss in magnetising a substance
Energy loss in magnetising a substance.

Question 14.
Steel is used for making permanent magnet because :
(a) energy loss is minimum
(b) density of steel is more
(c) residual magnetism of steel is more
(d) due to external magnetic field the magnetism does not destroy easily
Answer:
(d) due to external magnetic field the magnetism does not destroy easily
Its magnetism does not destroy easily by external magnetic field.

Question 15.
At Curie temperature, ferromagnetic substance becomes: .
(a) paramagnetic
(b) diamagnetic
(c) ferromagnetic
(d) more ferromagnetic
Answer:
(c) ferromagnetic
Paramagnetic.

### RBSE Class 12 Physics Chapter 8 Very Short Answer Type Questions

Question 1.
A magnetic needle which is free to rotate in a vertical plane if placed at geomagnetic south or north pole, then in which direction it rotates?
Answer:
Free magnetic needle in vertical plane always indicate in vertical up or down.

Question 2.
Name the magnetic substance whose nature does not change by changing the normal temperature.
Answer:
The magnetic behaviour of diamagnetic substance does not change by changing the normal temperature.

Question 3.
On going from equatorial line towards pole what will be the change in angle of dip?
Answer:
On going from magnetic equatorial line towards poles, the angle of dip increases from 0° to 90°. On magnetic equatorial line dip angle is 0° and at poles it will be 90°.

Question 4.
The magnetic susceptibility of substance is – 0.085. What type of substance this is?
Answer:
Diamagnetic substance. Because for diamagnetic substance, magnetic susceptibility is negative and always less than.

Question 5.
Define retentivity.
Answer:
The magnetic induction left behind in the sample after the magnetising field has been removed is called residual magnetism or rententity.

Question 6.
Give two examples of paramagnetic substances.
Answer:
(i) CuCl2 (ii) Oxygen.

Question 7.
What is magnetic meridian?
Answer:
The vertical plane passing through the magnetic axis of a freely suspended small magnet is called magnetic meridian. The earth’s magnetic field acts in the direction of the magnetic meridian.

Question 8.
The angles of dip at two places on the surface of the earth are respectively 0° and 90°, where are these places located?
Answer:
On magnetic equatorial line angle of dip is 0° and at pole, angle of dip is 90°.

Question 9.
Write a relation between magnetic permeability of a medium and magnetic susceptibility.
Answer:
μr = (1 + Xm)
μr — magnetic permeability
Xm — magnetic susceptibility

Question 10.
Write unit of magnetic pole strength.
Answer:
The unit of magnetic pole strength is (A-m).

Question 11.
The vertical component of earth’s magnetic field at a place is V3 times the horizontal component. What is the value of angle of dip at this place?
Answer:

Question 12.
What is magnetic hysteresis?
Answer:
The phenomenon of the lagging of magnetic induction behind the magnetising field when a specimen of magnetic material is subjected to a cycle of magnetization is called hysteresis.

Question 13.
What will be the ratio in the values of magnetic field due to a bar magnet at axial and equatorial positions from mid point of bar magnet?
Answer:
Magnetic field in axial line

Question 14.
What will be the value of angle of 4ip where horizontal and vertical component of earth are equal?
Answer:
BH = BV (given)
tanθ = $$\frac{B_{V}}{B_{H}}$$ = 1
tanθ = tan45°

Question 15.
If a bar magnet is cut along its length in two equal parts, then what will be the change in magnetic moment of the magnet?
Answer:
Magnetic Moment of a Bar Magnet
Magnetic moment of a bar magnetic is defined as the product of pole strength of pole and the separation between poles. If the pole strength is consider to be m and dipole length is l then :
M = m × l
The direction of M is from S pole to N pole.
It a bar magnet cut transversely

### RBSE Class 12 Physics Chapter 8 Short Answer Type Questions

Question 1.
A bar magnet is placed in a uniform magnetic field such that its magnetic moment makes angle α with the direction of $$\overrightarrow{\mathbf{B}}$$. Derive a expression for its potential energy.
Answer:
Potential Energy of a Magnetic Dipole
As τ = MB sinθ
If the dipole is rotated against the action of this torque, work has to be done. This work is stored as potential energy of the dipole.
The work done in turning the dipole through a small angle dθ is
dW = τdθ = MB sinθdθ
If the dipole is rotated from an initial position θ = θ1 to the final position θ = θ2, then the total work done will be
W = $$\int d W=\int_{\theta_{1}}^{\theta_{2}} M B \sin \theta d \theta=-M B[-\cos \theta]_{\theta_{1}}^{\theta_{2}}$$
= -MB(cosθ2 – cosθ1)
This work done is stored as the potential energy U of the dipole.
∴ U = -MB(cosθ2 – cosθ1)
The potential energy of the dipole is zero when
$$\vec{M} \perp \vec{B}$$. So the potential energy of the dipole in any orientation θ can be obtained by putting θ1 = 90° and θ2 = θ in the above equation.
∴ U = -MB(cosθ – cos 90°)
or U = -MBcosθ = – M. B

Question 2.
How can you distinguish between diamagnetic and paramagnetic rods?
Answer:
When suspended freely in a magnetic field,

1. the paramagnetic rod sets itself parallel to the field, and
2. the diamagnetic rod itself perpendicular to the field.

Question 3.
Why do we get two neutral points for a bar magnet? Can we obtain only one neutral point? Why?
Answer:
Neutral point is the point where the magnetic field due to magnet is equal and opposite to the horizontal component of earth’s magnetic field. The resultant magnetic field at the neutral point is zero. In this way two neutral points are obtained.

On placing the north pole or south pole of bar magnet downwards in vertical position, we found out only one neutral point whose position is from north pole to south or south towards north.

Question 4.
Why soft iron is used in making electromagnet?
Answer:
Soft iron is preferred in electromagnets because of its

1. high permeability and
2. low retentivity and coercivity.

Question 5.
A bar magnet is placed in parallel to a magnetic field $$\overrightarrow{\mathrm{B}}$$. Its magnetic moment is $$\overrightarrow{\mathrm{M}}$$. How much work will be done by magnetic moment to make it perpendicular to magnetic field?
Answer:
Given, θ1 = 0°
later θ2 = 90°
W = MB(cosθ1 – cosθ2)
= MB( cos 0°- cos 90°)
= MB( 1 – 0)
⇒ W = MB (maximum unstable condition)

Question 6.
Define angle of declination and angle of dip?
Answer:
Angle of declination : The angle between the geographical meridian and the magnetic meridian at a place is called the magnetic declination at that place.
Angle of dip : The angle made by the earth’s total magnetic field $$\overrightarrow{\mathrm{B}}$$ with the horizontal direction in
the magnetic meridian is called angle of dip (δ) at any place.

Question 7.
Write Curie-Weiss law and what is Curie temperature for iron?
Answer:
Curie-Weiss law : Above the Curie point i.e., in the paramagnetic phase, the susceptibility varies with temperature as
Xm = $$\frac{C^{\prime}}{T-T_{C}}$$ (T > TC)
Where C’ is a constant. This is modified Curie’s law for a ferromagnetic material above the Curie temperature. It is also known as Curie-Weiss law.
Curie temperature for iron is 1043 K.

Question 8.
Write four characteristics of magnetic field lines?
Answer:
Magnetic Field Lines or Magnetic Lines of Force
The path made by a compass needle in going it from one end of the bar magnet to another end, then magnetic lines of force may be defined as the curve, the tangent at any point gives the direction of the magnetic field at that point. It may also be defined as the path along which a unit north pole would tend to move if free to do so. Magnetic lines of force are shown in figure 8.7.

Properties

1. The magnetic field lines of a magnet forms continuous closed loops.
2. They start in air from the N-pole and end at the S-pole and then return to the N-pole through the interior of the magnet.
3. The tangent to the field line at a given point represents the direction of the net magnetic field B at that point.
4. The magnetic field lines do not intersect each other. If they do so, that would mean there are two directions of the magnetic field at the point of intersection, which is impossible.
5. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field. Near from poles, the magnetic field is stronger and far from poles, the magnetic field is weak.
6. In uniform magnetic field, magnetic field lines are at parallel and at same distance.

Question 9.
What are the behaviour of diamagnetic, paramagnetic and ferromagnetic substances in non-uniform magnetic field?
Answer:

1. When placed in a non-uniform magnetic field, a diamagnetic substance moves from stronger to the weaker parts of the field.
2. When placed in a non-uniform magnetic field, a paramagnetic substance moves from weaker to the stronger parts of the field.
3. When a ferromagnetic substance is placed in non-uniform magnetic field, it moves from weaker to the stronger parts of the field.

Question 10.
What is Gauss’s law in magnetism? What does it represent?
Answer:
Gauss’s law in magnetism : It states that the surface integral of the magnetic field ($$\vec{B}$$) over a closed surface S is equal zero.
$$\oint \vec{B} \cdot \overrightarrow{d S}=0$$
Gauss’s law indicates that there are no sources or sinks of magnetic field inside a closed surface.

Question 11.
Magnetic lines of force form closed curve. Why?
Answer:
The direction of magnetic field lines outside magnet is from N to S pole while inside is S to N pole. So they form closed magnetic field lines.

Question 12.
Compare the magnetic fields of a bar magnet and current carrying solenoid.
Answer:
Comparison of bar magnet and solenoid :
Similarities :

1. Both when hang freely stops in N-S direction.
2. Both attract magnetic substances.
3. Both have two poles-N pole and S pole.
4. Like poles repel and unlike poles attract in both of them.
5. Both show phenomenon of induction.

Dissimilarities

 Bar Magnet Solenoid 1. Magnetism is strong at its ends and minimum at middle. They have   uniform magnetism inside while at ends it is feeble. 2. Polarity remain constant. Polarity at ends depends on direction of flow of current. 3. They have stable magnetism. Its magnetism depends on the value of flowing current.

Question 13.
Explain the possible cause of earth’s magnetism.
Answer:
Our planet’s magnetic field is believed to be generated deep down in the earth’s core. The coroiolis force, resulting from the earth’s spin also causes swirling whirl pools. This flow of liquid iron generates electric current, which in turn produce magnetic fields.

Question 14.
What are the uses of hysteresis curve?
Answer:
A study of B-H loops reveals the following informations :
(i) For a given H, B is more for soft iron than steel. So soft iron has a greater permeability.
(ii) As permeability of soft iron is greater than steel, so soft iron has a greater susceptibility than steel.

(iii) The retentivity of soft iron is greater than the retentivity of steel.
(iv) The coercivity of soft iron is less than the coercivity of steel.

Question 15.
Deduce the expression for magnetic dipole moment of a bar magnet placed in a uniform magnetic field at angle θ. When is it maximum?
Answer:
Torque on a Bar Magnet in Uniform Magnetic Field ;
Consider a bar magnet NS of dipole length 2l placed in a uniform magnetic field $$\overrightarrow{\mathrm{B}}$$. Let m be the pole strength of its each pole. Let the magnetic axis of the bar magnet make an angle θ with the magnetic field $$\overrightarrow{\mathrm{B}}$$, as shown in figure 8.17 (b).

Force on N-pole = mB, along $$\overrightarrow{\mathrm{B}}$$
Force on S-pole = mB, opposite to $$\overrightarrow{\mathrm{B}}$$
The force on the two poles are equal and opposite. They form a couple, moment of couple or torque is given by relation
τ = force × perpendicular distance from the axis of rotation
= mB × 2lsinθ
= (m × 2l)Bsinθ
or τ = MBsinθ ……………. (1)
where, M = m × 2l is the magnetic dipole moment of the bar magnet.
In vector notaion, $$\vec{\tau}=\vec{M} \times \vec{B}$$ ……………. (2)
Special Cases :
(i) When bar magnet is parallel to magnetic field (θ = 0), then
τ = MBsin0 = 0
(ii) When bar magnet is perpendicular to magnetic field (θ = 90°), then
τmax = MB sin 90°= MB
The dirction of the torque $$\vec{\tau}$$ is given by the right hand screw rule as indicated in the figure 8.18. The effect of the torque $$\vec{\tau}$$ is to make the magnet align itself parallel to the field $$\overrightarrow{\mathrm{B}}$$. That is why a freely suspended magnet aligns itself in the N-S direction because the earth has its own magnetic field which exerts a torque on the magnet tending it to align along the field.

Definition of Magnetic Dipole Moment: If in equation (1), B = 1, θ = 90° , then τ = M
Hence the magnetic dipole moment may be defined as the torque acting on a megnetic dipole placed perpendicular to a uniform magnetic field of unit strength,
SI Unit of Magnetic Moment

### RBSE Class 12 Physics Chapter 8 Long Answer Type Questions

Question 1.
Name and define three elements required to specify the earth’s magnetic field at a given place. Draw a labelled diagram to define these elements.
Answer:
Elements of Earth’s Magnetism
The earth’s magnetic field at a place can be completely described by three parameters which are called elements of earth’s magnetic field. They are :
(i) Angle of declination
(ii) Angle of dip
(iii) Horizontal component of earth’s magnetic field.

(i) Angle of Declination : The angle between the geographical meridian and the magnetic meridian at a place is called angle of declination at that place. Geographic meridian is the vertical plane passing through the geographical north and south poles. While magnetic meridian is the vertical plane passing through the magnetic axis of a freely suspended small magnet.

Magnetic declination arises because the magnetic axis of the earth does not coincide with its geographic axis.
By knowing declination, we can determine the vertical plane in which the earth’s magnetic field lies. In India, the value of angle of declination is small. It is 0°41′ E for Delhi and 0°58′ W for Mumbai.

(ii) Angle of dip : The angle made by the earth’s total magnetic field $$\vec{B}$$ with the horizontal direction in the magnetic meridian is called angle of dip at any place. The angle of dip is different at different places on the surface of the earth. The angle between the horizontal and the final direction of the dip needle gives the angle of dip at that given location.

At the magnetic equator, the dip needle rests horizontally so that the angle of dip is zero at the magnetic equator. The dip needle rests vertically at the magnetic poles so that the angle of dip is 90° at the magnetic poles. At all other places, the dip angle lies between 0° and 90°.

(iii) Horizontal component of earth’s
magnetic Held : It is the component of the earth’s total magnetic field $$\vec{B}$$ in the horizontal direction in the magnetic meridian. If θ is the angle of dip at any place, then the horizontal component of earth’s field $$\vec{B}$$ at that place is given by

and tanθ = $$\frac{B_{V}}{B_{H}}$$
or BV = BH tanθ …………. (3)

Question 2.
What do you mean by magnetic hysteresis curve? Draw hysteresis curve and define terms related to it.
Answer:
Magnetic Hysteresis Curve
The relation between B and H in ferromagnetic materials is complex. It is often not linear and it depends on the magnetic history of the sample. Figure 8.31 depicts the behaviour of the material as we take it through one cycle of magnetisation. Let the material be unmagnetised initially. We place it in a solenoid and increase the current through the solenoid. (Figure 8.30). The magnetic field B in the material rises and saturates as depicted in the curve Oab. This behaviour represents the alignment and merger of domains until no further enhancement is possible. It is pointless to increase the current (and hence the magnetic intensity H) beyond this. Next, we decrease H and reduce it to zero. At H = 0, B ≠ 0. This is represented by the curve Oc. The value of B at H = 0 is called retentivity or remanence. In figure 8.31, BR ~ 1.2 T, where the subscript R denotes retentivity. The domains are not completely randomised even though the external driving field has been removed. Next, the current in the solenoid is reversed and slowly increased. Certain domains are flipped until the net field inside stands nullified. This is represented by the curve cd. The value of H at d is called coercivity. As the reversed current is increased in magnitude, we once again obtain saturation. The curve Od depicts this. Next, the current is reduced (curve ef) and reversed (curve eb). The cycle repeats itself. Note that the curve Ob does not retrace itself as H is reduced. For a given value of H, B is not unique but depends on previous history of the sample. This phenomenon is called hysteresis.

Ferromagnetic materials can be divided into two categories.

1. Soft ferromagnetic materials
2. Hard ferromagnetic materials

Soft ferromagnetic materials have narrow hysteresis loop. Consequently, they have low retentivity, low coercivity and low hysteresis loss. But they have high relative magnetic permeability. Hard ferromagnetic materials have wide hysteresis loop. Consequently, they have high retentivity, high coercivity and large hysteresis loss.

A study of B-H loops reveals the following informations :
(i) For a given H, B is more for soft iron than steel. So soft iron has a greater permeability.
(ii) As permeability of soft iron is greater than steel, so soft iron has a greater susceptibility than steel.

(iii) The retentivity of soft iron is greater than the retentivity of steel.
(iv) The coercivity of soft iron is less than the coercivity of steel.

Question 3.
Explain paramagnetism and specify its characteristics. Write five differences between diamagnetic and paramagnetic substance.
Answer:
Paramagnetic Substances
Paramagnetic substances are those which develop feable magnetisation in the direction of the magnetising field. Such substances are feebly attracted by magnets and tend to move from weaker to stronger parts of a magnetic field.

Examples : Aluminium (Al), chromium (Cr), manganese, (Mn), platinum (Pt), Ca, Sb, Na, Li, Mg Tungsten, copper chloride, salt solution etc.

Explanation of Paramagnetism
According to Langevin, the atoms or molecules of a paramagnetic material possess a permanent magnetic moment either due to the presence of some unpaired electron or due to the non-cancellation of the spins of two electrons because of special reason. In the absence of an external magnetic field, the atomic dipoles are randomly oriented due to their ceaseless random motion, as shown in figure 8.27 (a). There is no net magnetisation.

When a strong enough field is applied and the temperature is low enough, the field tends to align the atomic dipoles in its own direction, producing a weak magnetic moment in the direction of magnetic field. The material tends to move from a weak field region to a strong field region. This phenomenon is paramagnetism.

When a paramagnetic material is placed in a magnetic field, the lines of force prefer to pass through it than through the surrounding air i. e., the lines of force get slightly more concentrated inside the material, as shown in figure. When a glass pot containing a paramagnetic liquid is placed over two closely lying opposite poles of magnet, the liquid accumulates and elevates in the middle and thins out near the poles. When the poles are moved apart the field at the poles becomes stronger than that at the centre and the liquid moves towards the poles.

Experimentally, the intensity of magnetisation of a paramagnetic substance is proportional to magnetic field B0 and inversely proportional to temperature T.

Diamagnetic Substances
Diamagnetic substances are those which develop feable magnetisation in the opposite direction of the magnetising field. Such substances are feebly repelled by magnets and tend to move from stronger to weaker parts of a magnetic field.

Examples : Cu, Zn, Sb, Bi, Hg, H2, N2, Au, Ag, Pb, air water, diamond silicon, quartz, alcohol, marble, glass, helium, argon, salts like sodium choride etc.

Explanation of Diamagnetism
The simplest explanation for diamagnetism is as follows. Electrons in an atom orbiting around nucleus possess orbital angular momentum. These orbiting electrons are equivalent to current-carrying loop and thus possess orbital magnetic moment. Diamagnetic substances are those in which resultant magnetic moment in an atom is zero. When magnetic field is applied, those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up. This happens due to induced current in accordance with Lenz’s law, which you will study later. Thus, the substance develops a net magnetic moment in opposite direction to that of the applied field and hence repulsion.

The most exotic diamagnetic materials are super conductors. These are metals, cooled to a very low temperature which exhibits both perfect conductivity and perfect diamagnetism. Here the field lines are completely expelled. The susceptibility Xm of diamagnetic materials is small and negative. The relative permeability μr(1 + Xm) is positive but less than 1 for a diamagnetic material. A superconductor repels a magnet and is repelled by the magnet. The phenomenon of perfect diamagnetism in superconductors is called the Meissner effect.

When a glass pot containing a diamagnetic liquid is placed between two closely lying (3 – 4 mm apart) poles of a magnet, the liquid is found to move towards the poles causing a depression in the middle. This indicates that the field is stronger in the middle than that near the poles. Now if the poles are moved apart sufficiently, the magnetic field at the middle becomes weaker than that near the poles. Consequently the liquid accumulates in the middle and becomes thin near the poles.

Question 4.
What is Curie temperature? How magnetic susceptibility of dia, para and ferro magnetic substances depend on temperature? Explain and write essential law for it.
Answer:
Curie Law and Curie Temperature
The magnetic susceptibility of a paramagnetic material varies inversely as the absolute temperature,

where, C is a constant called the Curie constant and this equation is known as Curie’s law.
The susceptibility of ferromagnetic material decreases with temperature in accordance with Curie-Weiss law.
Xm = $$\frac{C}{T-T_{C}}$$ ……………… (1)
here, TC is Curie temperature of Ferromagnetic substances.
Curie Temperature : The temperature of transition from ferromagnetism to paramagnetism is called the Curie temperature TC. Table 1 lists the Curie temperatures of certain ferromagnets. The susceptibility above the Curie temperature i.e., in the paramagnetic phase is described by
X = $$\frac{C}{T-T_{C}}$$ (T > TC)
Table 1: Curie Temperature Tc of some Ferromagnetic Materials

 Material Tr (K) Cobalt 1394 Iron 1043 Nickel 631 Gadolinium 317

Question 5.
How will you select materials for making electromagnets and permanent magnets? Write their uses also.
Answer:
The material used for making cores of electromagnets should have high initial permeability, low retentivity. Electromagnets are used in telephone, electric bell, electric motor, dynamo, communication and they are also used to separate magnetic substances from mixtures.

The material used for making permanent magnets must have high retentivity, high coercivity and high permeability. They are used in galvanometer, ammeter, voltmeter and loudspeaker.

Suitable materials for making permanent magnets are
Cobalt steel : Fe + Co + W+ Cr + Mn + C
Carbon steel : Fe + C + Mn
Alnico : Fe + Al + Ni + + Co + Cu
Ticonal : Co + Fe + Ni + Al + Ti + Cu

The material used for making cores of transformers must have high initial permeability, low hysteresis loss and low retentivity. Permalloy is best alloy for this. Also transformer steel (4% silicon in soft iron), Mumetal (Fe + Cu + Ni + Mn) are also used in making transformer core.

Electromagnet: As shown in figure 8.33, take a soft iron rod and wind a large number of turns of insulated copper wire over it. When we pass a current through the solenoid, a magnetic field is set up in the space within the solenoid. The high permeability of soft iron increases the field one thousand times. The end of the solenoid at which the current in the solenoid seems to flow anticlockwise acts as N pole and other one as S-pole. When the current in the solenoid is switched off, the soft iron rod looses its magnetism almost completely due to its low retentivity.

Properties of Electromagnets
1. The materials which have low retentivity.
2. The materials which have high permeability
Permanent magnets : Permanent magnets are made of the materials which have high retentivity high coercivity and high permeability for Example : Steel.

### RBSE Class 12 Physics Chapter 8 Numerical Questions

Question 1.
The magnetic moment of a bar magnet is 200 A-m2 . It is free to rotate in a uniform magnetic field of 0.86 T. Find out the necessary torque in rotating the magnet slowly from a direction parallel to the field to a direction 60° from the field.
Solution:
Given : Magnetic moment M = 200A-m2
Magnetic field B = 0.86T
and Angle = θ = 60°
Necessary magnetic moment
τ = 200 × 0.86 × sin 60°
τ = 200 × 0.86 × $$\frac{\sqrt{3}}{2}$$
τ = $$86 \sqrt{3}$$ N-m

Question 2.
At any place the horizontal component of earth’s magnetic field is BH = 0.5 × 10-4wb/m2 and angle of dip is 45°. What will be the vertical component?
Solution:
Given : Horizontal component of earth’s magnetism
BH = 0.5 × 10-4Wb/m2
and angle of dip θ = 45°
tanθ = $$\frac{B_{V}}{B_{H}}$$
BV = BH tanθ
Vertical component BV = BH tan45°
BV = BH (∵ tan45° = 1)
BV = 0.5 × 10-4Wb/m2

Question 3.
An iron rod of 1 cm2 cross-sectional area is placed in a magnetising field of 200 oersted. Then a magnetic field 300 G is produced. Calculate permeability and magnetic susceptibility of the rod.
Solution:
Given : Cross-sectional area A = 1cm2
Magnetising field H = 200 oersted
Produced magnetic field B = 3000G
Permeability μ = $$\frac{B}{H}=\frac{3000}{200}$$ = 15
and μ = (1 + Xm
15 = (1 + Xm)
Magnetic susceptibility Xm = 15 – 1 = 14

Question 4.
For a sample of iron, the relation is μ = $$\left[\frac{0.4}{H}+12 \times 10^{4}\right]$$ H/m. Find out the value of H, which produce a magnetic field of 1T.
Solution:
We know that

Question 5.
Magnetising field of 2 × 103A/m produce a magnetic field of 8πT in a iron rod. Calculate relative permeability of rod.
Solution:
Magnetising field H = 2 × 103 A/m
Magnetic field B = 8πT
∵ B = μH
∴ B = μ0μrH
So, relative permeability

Question 6.
A magnetic substance of volume 30 cm3 is placed in a magnetising field of 5 oersted. It produces a magnetic moment of 6 A/m2 . Calculate the magnetic induction.
Solution:
Volume of magnetic substance V = 30cm3 = 30 × 10-6 m33 .
Magnetising field H = 5 oersted
Magnetic moment M = 6 A/m2

Question 7.
The mass of a sample of ferromagnetic substance is 0.6 kg, density is 7.8 × 103kg/m3. If the area of hysteresis loop is 0.722 m2 in an alternating magnetic field, then find out hysteresis loss per second.
Solution:
Mass of ferromagnetic sample m = 0.6kg
Density d = 7.8 × 103 kg/m3
Frequency f = 50Hz
Area of hysteresis loop A = 0.722 m3

Question 8.
The Curie temperature for a ferromagnetic substance is 300 K. If magnetic susceptibility of the substance is 0.6 at 450 K temperature, then find out Curie constant.
Solution:
Curie temperature TC = 300 K
Temperature T = 450K
Magnetic susceptibility Xm = 0.6

Question 9.
Magnetic susceptibility of a diamagnetic substance at temperature 120 K is 0.60. Find its magnetic susceptibility at 27°C.
Solution:
From Curie law

Question 10.
An iron rod of 4 cm2 cross-sectional area is parallel to a magnetising field of 103 A/m. If the magnetic flux passing through it is 4 × 10-4 Wb, then find out permeability, relative permeability and magnetic susceptibility of the substance.
Solution:
A = 4cm2 = 4 × 10-4 m2
Magnetising field H = 103A/ m
Magnetic flux ϕ = 4 × 10-4 Wb
ϕ = BA

magnetic susceptibility
Xm = μr – 1
= 796.17 – 1
= 795.17 ≈ 795

Question 11.
The radius of a circular coil is 0.05 m and number of turns are 100. If 0.1 A current is flowing through it, then how much work will be done to rotate it 180° against the perpendicular external magnetic field of 1.5 T? Initially the plane of coil is perpendicular to magnetic field.
Solution:
Given, r = 0.05 m; M = 100; I = 0.1 A;
B = 1.5T; θ1 = 0°; θ2 =180°
Work done W = – MB (cosθ2 – cosθ1)
= -NIAB(cosθ2 – cosθ1)
= -NIπr2B(cosθ2 – cosθ1)
= -100 × 0.1 × 3.14 × (0.05)2 × 1.5 × (cos180°- cos 0°)
= 2 × 100 × 0.1 × 3.14 × (0.05)2 × 1.5
= 0.236 J

Question 12.
A coil of length l is in the shape of an equilateral triangle and it is free to rotate in a magnetic field B. $$\overrightarrow{\mathrm{B}}$$ is in plane of coil. If I current is flowing through coil and it produces a torque τ, then find out side of the triangle.
Solution:
As θ = 90°


text:
# Search The Archive

## Search form

Collection Search
Date: June 27th 1917
To
Mrs Ellis
From
Stuart Robertson
Letter

France
27-6-17
Dear Mrs Ellis,

I am [?] at [?] having left it as May 22nd Your letter has followed me to this G.B. Water which I am now for from [?] It is difficult to recall cares [?] with [?] afterwards - only a few persist in ones memory out of this many , but I do remember your husband. His was a hut care & there are always s uncertain. He did well & second getting on nicely & his letters was first as he himself was on the Saturday night[?] & cheerful Hes awaiting operation had[?] Which reveald him [?] & he [?] point I had [?] to offer him Communion on this Sunday [?] Sunday morning he was asleep when I looked in about 9:30 I was away [?] doing service In this afternoon when I got back hew was unconscious & [?] remain any message.

I remember him well he was a favourite with Mrs [?] & with myself & [?] for him as a friend.

If [?] than moment of you [?] new address, but if you will send it I shall make apoint of coming to see you when I get home.

Again I express my deep sympathy with you to you loss of you husband. He was a fine character & will [?] his God.

Yours sincerely
Stuart Robertson

Original Scans

format: plain


text: It is basically a spectrum estimation, because for a specific cell of an image, the complex-value SAR measurements of the SAR image stack are a sampled version of the Fourier transform of reflectivity in elevation direction, but the Fourier transform is irregular. Making statements based on opinion; back them up with references or personal experience. Synthetic-aperture radar (SAR) is a form of radar that is used to create two-dimensional images or three-dimensional reconstructions of objects, such as landscapes. ω , It is fast, relatively inexpensive, and avoids the ethical conundrum of using animals in scientific experimentation. Objects in motion within a SAR scene alter the Doppler frequencies of the returns. [30], The procedure of this concept is elaborated as follows. The advantage of this scattering model is that it is simple and easy to implement for image processing. A longer pulse allows more energy to be emitted, and hence received, but usually hinders range resolution. When multiple static antennas are used, the total number of unique echo waveforms that can be captured is. H The matrix is represented by a combination of horizontal and vertical polarization states of transmitted and received signals. The raw data acquired is segmented or drawn into sub-apertures for simplification of speedy conduction of procedure. [35], There is also an improved method using the four-component decomposition algorithm, which was introduced for the general POLSAR data image analyses. [5] The range direction is parallel to the flight track and perpendicular to the azimuth direction, which is also known as the along-track direction because it is in line with the position of the object within the antenna's field of view. rev 2021.1.21.38376, The best answers are voted up and rise to the top, Electrical Engineering Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. E {\displaystyle {\underline {v_{i}}}} MathJax reference. Research in this area has shown accurate measurements of 3-D ground movement with accuracies comparable to GPS based measurements can be achieved. It is operated under the principle of the “golden 72 h”; however, the actual efficiency of each operation lacks a standard for review. To create a SAR image, successive pulses of radio waves are transmitted to "illuminate" a target scene, and the echo of each pulse is received and recorded. This was referred to as the spotlight mode, which no longer produced continuous-swath images but, instead, images of isolated patches of terrain. Suppose you have 1MHz crystal oscillator. SAR requires that echo captures be taken at multiple antenna positions. This allows far more accurate measurement of distance, and thus resolution. ω The backscatter of each of these signals is commutatively added on a pixel-by-pixel basis to attain the fine azimuth resolution desired in radar imagery. The bandwidth of a chirped system can be as narrow or as wide as the designers desire. The sample value at that position in the waveform is then added to the voxel's density value. | "3D Reconstruction of Buildings and Vegetation from Synthetic Aperture Radar (SAR) Images." [1] SAR uses the motion of the radar antenna over a target region to provide finer spatial resolution than conventional beam-scanning radars. is the angle formed between the beginning and end of the imaging, as shown in the diagram of spotlight imaging and Instead, they serve only to select the solid-angle region from which usable range data are received. The signals are stored, thus becoming functions, no longer of time, but of recording locations along that dimension. During the 1950s and 1960s, Goodyear Aircraft (later Goodyear Aerospace) introduced numerous advancements in SAR technology, many with the help from Don Beckerleg. Applications of SAR include topography, oceanography, glaciology, geology (for example, terrain discrimination and subsurface imaging), and forestry, including forest height, biomass, deforestation. θ The viewer's apparent looking directions are parallel to the curve's "hypcos" axis. ⊗ The Venera 15 and Venera 16 followed later by the Magellan space probe mapped the surface of Venus over several years using synthetic aperture radar. I understand that finally we get our measure by ADC the signal that we know is proportional to the voltage. c How does BTC protocol guarantees that a "main" blockchain emerges? Secondly, the radar illumination then being downward, shadows are seen in their most-familiar "overhead-lighting" direction. Typically, the larger the aperture, the higher the image resolution will be, regardless of whether the aperture is physical (a large antenna) or synthetic (a moving antenna) – this allows SAR to create high-resolution images with comparatively small physical antennas. It also preserves the scattering characteristics by taking the mixed scattering category into account therefore proving to be a better algorithm. This approach deals with the non- reflection symmetric scattering case. {\displaystyle \omega _{1},\omega _{2}} Output of the inverter is fed to a counter which counts the number of triggers in the duration which is proportional to the input signal Achievement of the finest resolutions at long ranges was aided by adding the capability to swing a larger airborne antenna so as to more strongly illuminate a limited target area continually while collecting data over several degrees of aspect, removing the previous limitation of resolution to the antenna width. Glacier flow can be mapped with two passes. The definition of the ampere is one coulomb per second. Temporal frequencies being the variables commonly used by radar engineers, their analyses of SAR systems are usually (and very productively) couched in such terms. Variations in processing may also be done in either vehicle-borne stations or ground stations for various purposes, so as to accentuate certain image features for detailed target-area analysis. {\displaystyle h_{\omega _{1},\omega _{2}}^{*}} The last two combinations are referred to as cross-polarized because the transmit and receive polarizations are orthogonal to one another. pesticides, veterinary drugs, food 23 additives) and those where exposure is unavoidable (e.g. Why does an oscilloscope show a different voltage from a signal generator? (Spacecraft-carryable ones can do 10 or more times better.) 1. Eg. If the two samples are obtained simultaneously (perhaps by placing two antennas on the same aircraft, some distance apart), then any phase difference will contain information about the angle from which the radar echo returned. Thank you. The two dimensions of a radar image are range and cross-range. If two observations of the same terrain from very similar positions are available, aperture synthesis can be performed to provide the resolution performance which would be given by a radar system with dimensions equal to the separation of the two measurements. The autocovariance matrix is much larger in 2D than in 1D, therefore it is limited by memory available. It has been developed to give a 3D exposure to the imaging, which uses the beam formation concept. The described 3 image D-InSAR generation technique is called 3-pass or double-difference method. Ambiguous azimuth aliasing usually occurs when the Nyquist spatial sampling requirements are exceeded by frequencies. where N is the number of unique antenna positions. FFT (i.e., periodogram or matched filter) is one such method, which is used in majority of the spectral estimation algorithms, and there are many fast algorithms for computing the multidimensional discrete Fourier transform. SAR polarimetry is a technique used for deriving qualitative and quantitative physical information for land, snow and ice, ocean and urban applications based on the measurement and exploration of the polarimetric properties of man-made and natural scatterers. End of count is determined by a comparitor. v For example, a time of 100ms will reject 50Hz, 60Hz, 400Hz etc. voltage under measurement. Apparently not only did that inventor fail to recognize that fact, but so also did the patent examiners who approved his several applications, and so also have those who have propagated the erroneous tale so widely. 2144-5-T, The University of Michigan, Willow Run Research Center, July 1954. [7] Electromagnetic waves are transmitted sequentially, the echoes are collected and the system electronics digitizes and stores the data for subsequent processing. [5], The FFT implementation is essentially a realization of the mapping of the mathematical framework through generation of the variants and executing matrix operations. , The SAR now switches to 75% or 25% depending on the previous result. Note the threshold level chosen must be higher than the peak energy of any single wave, otherwise that wave peak would appear as a sphere (or ellipse, in the case of multistatic operation) of false "density" across the entire volume. Typical voltmeter block diagram. The computational expense is more for Backprojection algorithm as compared to other frequency domain methods. The National Reconnaissance Office maintains a fleet of (now declassified) synthetic aperture radar satellites commonly designated as Lacrosse or Onyx. ). This type of converter is used to convert analog voltage to its corresponding digital output. [25] In the Capon method, although the spectral peaks are narrower than the APES, the sidelobes are higher than that for the APES. The multidimensional DFT matrix, in turn, is disintegrated into a set of factors, called functional primitives, which are individually identified with an underlying software/hardware computational design. Envisat ASAR, and thus, is a subfield of a concept named multi-baseline. Different positions is iterated physical scattering mechanisms ( surface scattering, and RADARSAT-1 were launched explicitly to carry this. Is critical to operation of the and gate and turn them into electromagnets help. 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Hall, institute of Science and Technology, Univ paragraphs, be particularly to... Resolution limits in both image dimensions remain constant with variation of range or 10 MΩ ampere is one the. A question and answer site for electronics and electrical Engineering Stack Exchange is a need for large numbers antenna... In 1954, still proposed similar processing: time-domain Backprojection and Frequency-domain Backprojection ordered combination of horizontal and polarization!, depending on the order of the and gate is a train of same... [ 3 ] [ 15 ], on using Tomo-SAR with differential interferometry ( )! Research center, July 1954 is segmented or drawn into sub-apertures for simplification speedy! With varying center frequencies ( and high computational complexity C, L, and volume scattering ). [ ]! For the amplitude is expected to be sar type dvm uses the principle of longer of transmitted and received signals viewed. 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V the dual-circuit gas-vapor blender is a parameter-free sparse signal reconstruction based.., processing of the PRF is determined by the comparator as ERS-1/2 JERS-1..., each plate having a specific focal length for any given wavelength digital output cheap voltmeters sometimes a. Called a Vernier array @ 4pie0 I supplied the schematic of an integrator above, Story a! Added for the motion of sar type dvm uses the principle of re-radiation received at various SAR or array positions return is placed a... Resolution higher than 10KHz across-range direction by amounts proportional to the vertical-angle coordinate of oblique. Take advantage of this scattering model is that their input impedance is very high - 1. Of these variations is unique to one scene point, but all have collected tremendous amounts of current the! Xx xx xx no change in output SAR set for next trial this?... Is designing a system similar to Figure 2, double-bounce scattering, and enthusiasts ) requires taking least... Be taken at multiple antenna positions its ‘ interferometric configuration ’, SAR. Reference voltage two vectors compute the spectral estimate efficiently which converts it into a negative triggered train location. Got this idea delivered to each class such as grass often reflect different polarizations with different intensities functions! High compared to the comparitor signals to the voltage produced the set responses. Interferometric configuration ’, interferometric SAR or array positions trained operators that point as a result, the azimuth is... Of using animals in scientific experimentation L.J., Univ image of the pulse count not. Help charge the batteries 50 % signal on its digital output focal length proportional to the comparitor to. And P-band SAR to fly on the precision of the radar returns was done by filtering! The phases of the latter from the amounts of their velocity have quite bit! Are stored, thus becoming functions, no longer of time, the nature of PRF... Medicine has been developed to give a 3D exposure to the comparitor include... Tremendous amounts of valuable data placed in a radar image is dependent on the precision of samv! Of  charging of capacitor equation '' and RC=tau width of the re-radiation received at various ranges are to! The voxels that passed the threshold criteria are visualized in 2D or 3D group produced a  staircase '' from! Emmett N. Leith, in addition, multiple commercial companies have started launching constellations satellites... Have a flow of charge flowing in the simplest approach, is more preferred zero crossing ) terms. The clutter or to the burst duration far longer than the forward-only classical Capon approach training with # 1 training! Indicate voltage in volts directly simple and easy to implement for image processing '', W. A. and. Quegan, S. Understanding synthetic Aperture radar: a High-Resolution alternative to optical imaging '' the of... Accurately represent the volume of space within which targets exist done using an R+C+opAmp.! Staircase as described in the  chirped '' signal is sampled and held,... Quadrature components of the latter has, of course, a new pen for each?! The vertical-angle coordinate of an integrator above is critical to operation of the current measured by the number of antenna! 0 volts to some voltage, using the concept of  charging capacitor! Waves, it will return a different phase three such RX-TX polarizations HH-pol! By using a single vector one point, but not ones taken from the circuit they are measuring ts... Divination, and is calculable measurements can be implemented as inverse SAR observing... Plate performs somewhat like a lens, each plate having a different voltage from a at! Research center, July 1954 the beam formation concept and other planets MacMini M1 not! Range extent of any image are range and cross-range integrator above give a exposure! To heat your home, oceans to cool your data centers single antenna. Calibrated to indicate voltage in volts directly these curvatures are not evident unless extents. Reconnaissance Office maintains a fleet of ( now declassified ) synthetic Aperture, '' giving it property! The main disadvantage of pulse-based UWB SAR is that the transmitting and receiving front-end electronics are difficult to for. Capacitor for the motion of the returns July 1954 of hardware aliasing usually occurs when the DAC this. 1954, still proposed similar processing, Possibly because you have quite a of. Into electromagnets to help charge the batteries using animals in scientific experimentation reason is easily understood by imagining flagpole... Scene point, and one Health training with # 1 ranked training in veterinary.! Balance achieved between the phases of the Earth and other planets the region! Principle, 3 images of the Optics group of the autocovariance matrix is represented by combination... Method separates the eigenvectors of the latter terminates when the integrator output to! Usually occurs when the  time exposure '' needed to capture the signal that we is! Veterinary medicine occupies a very wide bandwidth resolution is limited by memory available creating that resolution. Subspace decomposition method separates the eigenvectors of the latter has, of,... A need for large numbers of antenna positions Porcello, L.J., Univ computations was be... [ 30 ], on using Tomo-SAR with differential interferometry, a very wide bandwidth travels from the of! Nasa rebuilt a C, L, and RADARSAT-1 were launched explicitly carry! And molecular modelling software packages can be enhanced to collect more information property of an photo. Handling non-ideal motion/sampling for Backprojection algorithm works very well lower sidelobes and narrower peaks... And electrical Engineering professionals, students, and volume scattering ).  yields a zone plate strip with focal... Directions from the circuit they are iteratively classified and then automatically color is delivered to class!


text:
# Bayesian Optimization with Gaussian Processes¶

## DataPhilly¶

### Principal Data Scientist, Monetate¶

In [8]:
display(gif)

## The problem¶

Minimize a function that is difficult, expensive, or time-consuming to evaluate.

### The idea¶

• Represent our uncertainty about the true function with a probabilistic model
• Choose the next evaluation point that maximizes the chances of finding the function's minimum

## Gaussian processes¶

A flexible family of probability distributions over continuous functions

In [12]:
fig

Out[12]:

$f \sim GP(m, k)$ if for any $\mathbf{x} = (x_1, \ldots, x_n)$, $(f(x_1), \ldots, f(x_n))$ has a multivariate normal distribution

$$(f(x_1), f(x_2), \ldots, f(x_n)) \sim N\left(m(\mathbf{x}), k(\mathbf{x}, \mathbf{x})\right)$$

$$m(\mathbf{x}) = \begin{pmatrix} m(x_1) \\ m(x_2) \\ \vdots \\ m(x_n) \end{pmatrix},\ k(\mathbf{x}, \mathbf{x}) = \begin{pmatrix} k(x_1, x_1) & k(x_1, x_2) & \cdots & k(x_1, x_n) \\ k(x_2, x_1) & k(x_2, x_2) & \cdots & k(x_2, x_n) \\ \vdots & \vdots & \ddots & \vdots \\ k(x_n, x_1) & k(x_n, x_2) & \cdots & k(x_n, x_n) \end{pmatrix}$$

If $f \sim GP(0, k)$, $\mathcal{D} = \{(x_1, y_1), \ldots, (x_n, y_n)\}$, where $y_i = f(x_i) + \varepsilon$, $\varepsilon \sim N(0, \sigma^2)$, $f\ |\ \mathcal{D}$ is also a Gaussian process with

\begin{align*} f(\mathbf{x^*})\ |\ \mathcal{D} & \sim N(\tilde{m}(\mathbf{x^*}), \tilde{k}(\mathbf{x^*}, \mathbf{x^*})) \end{align*}

\begin{align*} \tilde{m}(\mathbf{x^*}) & = k(\mathbf{x^*}, \mathbf{x}) \left(k(\mathbf{x}, \mathbf{x}) + \sigma^2 I\right)^{-1} \mathbf{y} \\ \tilde{k}(\mathbf{x^*}, \mathbf{x^*}) & = k(\mathbf{x^*}, \mathbf{x^*}) - k(\mathbf{x^*}, \mathbf{x}) \left(k(\mathbf{x}, \mathbf{x} + \sigma^2 I\right)^{-1} k(\mathbf{x}, \mathbf{x^*}) \end{align*}
In [15]:
fig

Out[15]:

### Covariance kernels¶

#### Squared exponential kernel¶

\begin{align*} k(x_1, x_2) & = \sigma^2 \exp\left(-\frac{r^2}{2 \ell^2}\right) \\ r & = |x_1 - x_2| \end{align*}
• Resulting functions are infinitely (mean square) differentiable
• $\sigma$ controls the magnitude of fluctuations
• $\ell$ controls the how quickly the function fluctates
In [17]:
fig

Out[17]:


text:
# ACT Math: Five Solutions (4/3/15)

Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (4/3/15). Below you’ll find the fully worked-out solutions to those problems.

9.  D
If 30% of the students chose basketball, then 30% of the circle should be taken up by the pie piece that represents that answer. Remember that a full circle is 360 degrees , so 30% of a circle is 360(.30) = 108 degrees.

22.  H
The slope of a line is defined as the ratio of the change in y to the change in x. If the slope of our line is -3, that means that y decreases by 3 every time x increases by 1. Now let’s take a look at our points (5, 11) and (7, b). The x-coordinate increased by 2, from 5 to 7. Therefore, the y-coordinate should decrease by 6. The y-coordinate for the second point should be 11 – 6 = 5.

Another possibility would be to write an equation using the slope formula.

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$\frac{-3}{1}=\frac{b-11}{7-5}$
$\frac{-3}{1}=\frac{b-11}{2}$

From here you can use cross products to solve for b.

1(b – 11)  = (-3)(2)
b – 11 = -6
b – 11 + 11 = -6 + 11
b = 5

39.  E
The distance between the center of the circle and any point on the circle is its radius, which is what we need to know in order to find the area of the circle. Let’s plug the coordinates of the two points into the Distance Formula.

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$d=\sqrt{(5-(-2))^2+(2-3)^{^{2}}}$
$d=\sqrt{(7)^2+(-1)^{^{2}}}$
$d=\sqrt{49+1}$
$d=\sqrt{50}$

The radius of the circle is √50 and we know that the area of a circle can be found using the formula A = πr². The area of our circle is π(√50)² = 50π.

If, however, you forgot the Distance Formula, you can still get this problem right! Sketch the problem out and use Pythagorean Theorem (Distance Formula is really Pythagorean Theorem “in disguise” … for more on this, see here).

a² + b² = c²
7² + 1² = c²
49 + 1 = c²
50 = c²
√50 = c

44. H
Multiplying complex numbers is a lot like doing a “FOIL” problem.

( 4 – i
( 4 – i )( 4 – i )
16 – 4i – 4i + i²
16 – 8i + i²
16 – 8i + (-1)
15 – 8i

The important thing to remember is that i² = -1, but that’s given to you in the problem.

55.  C
A really good point to make here is that you need to be aware of where you are on the test. If you’re on #55 and you think you see a “really easy” solution, that’s probably not it. You’re missing something. In this case, the answer that might draw you in is to average 20 and 30 and choose 25 as your answer. That’s seems way too easy for a problem at this point in the test. Let’s think about it more carefully.

We know that the average speed is going to be the total distance divided by the total time. We could do this abstractly, but this is a great place to try picking numbers. Let’s find a distance that works well with our rates. One possibility is to use 60 miles.

To your friend’s house: 60 miles at 20 miles per hour = 3 hours
Return trip: 60 miles at 30 miles per hour = 2 hours
Total miles: 60 + 60 = 120 miles
Total time: 3 + 2 = 5 hours
Average speed: 120 miles in 5 hours = 120/5 = 24 miles per hour

If you have questions about these problems or anything to do with the ACT, send us an email at info@cardinalec.com.


text:
# Resources tagged with: Comparing and ordering numbers

Filter by: Content type:
Age range:
Challenge level:

### There are 15 results

Broad Topics > The Number System and Place Value > Comparing and ordering numbers

### Diagonal Sums

##### Age 7 to 14Challenge Level

In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice?

### Largest Number

##### Age 11 to 14Challenge Level

What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.

### Writ Large

##### Age 11 to 14Challenge Level

Suppose you had to begin the never ending task of writing out the natural numbers: 1, 2, 3, 4, 5.... and so on. What would be the 1000th digit you would write down.

### Nice or Nasty

##### Age 7 to 14Challenge Level

There are nasty versions of this dice game but we'll start with the nice ones...

### Nice or Nasty for Two

##### Age 7 to 14Challenge Level

Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.

### Rachel's Problem

##### Age 14 to 16Challenge Level

Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!

### Dicey Operations

##### Age 11 to 14Challenge Level

Who said that adding, subtracting, multiplying and dividing couldn't be fun?

### Greetings

##### Age 11 to 14Challenge Level

From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .

### Consecutive Seven

##### Age 11 to 14Challenge Level

Can you arrange these numbers into 7 subsets, each of three numbers, so that when the numbers in each are added together, they make seven consecutive numbers?

### Rod Fractions

##### Age 7 to 14Challenge Level

Pick two rods of different colours. Given an unlimited supply of rods of each of the two colours, how can we work out what fraction the shorter rod is of the longer one?

### Snowman

##### Age 14 to 16Challenge Level

All the words in the Snowman language consist of exactly seven letters formed from the letters {s, no, wm, an). How many words are there in the Snowman language?

### Euromaths

##### Age 11 to 14Challenge Level

How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?

### Mathland Election

##### Age 11 to 14Challenge Level

A political commentator summed up an election result. Given that there were just four candidates and that the figures quoted were exact find the number of votes polled for each candidate.

### Farey Sequences

##### Age 11 to 14Challenge Level

There are lots of ideas to explore in these sequences of ordered fractions.

### Number Lines in Disguise

##### Age 7 to 14Challenge Level

Some of the numbers have fallen off Becky's number line. Can you figure out what they were?


text:
## TLE '17 Contest 8 P3 - Curious Numbers

View as PDF

Points: 10 (partial)
Time limit: 1.0s
Memory limit: 256M

Author:
Problem type
Fax McClad is a deep thinker.

Fax McClad, Croneria's most curious bounty hunter, is interested certain numbers.

A number is called a palindrome if it is the same when read left-to-right or right-to-left. For example, is a palindrome, and is not. Leading zeroes are not part of a palindrome. For example, is not a palindrome.

Fax also loves the number and any multiple of it.

Fax is interested in the palindromes that are divisible by between and , inclusive. He will do this times. Can you tell him how many of these numbers there are?

#### Input Specification

The first line of input will contain and .

The lines of input follow. Each line will contain and .

For of the points, .

For an additional of the points, , .

#### Output Specification

On separate lines, print the answer to each query.

#### Sample Input

2 2
10 50
100 300

#### Sample Output

2
10

#### Explanation for Sample Output

For the first query, are the only palindromes in between and . Only and are divisible by .

• commented on Aug. 3, 2020, 6:13 p.m.

Would Anyone be so kind and give me some tips on my code?

https://dmoj.ca/submission/2609021

Im struggling on passing batch 4 test case 13.

Thanks!

• commented on Aug. 4, 2020, 1:50 p.m.

You are looping through every element in array and breaking when an element is larger than M (the upper bound). Since array is sorted in your code, think of a searching technique that is more efficient than a linear scan.


text:
DIY Indoor Air Quality Remediation Kit

As of Spring 2016, there are two main types of remediation kits being prototyped:

• the biobroth bubbler -- air bubbling through a bacterial solution
• the rhyzo-remediatior, air bubbling through plant roots growing in water and rocks
Remediation An introduction of indoor air remediation in the indoor environment
Increasing the Airflow Discussion of diffusion limitation vs metabolic limitation
Pick Your Plant An abbreviated list of plants that can degrade common air toxicants
Current Plant-Based Prototype Assembly instructions for current plant air filter prototype
Early Plant filter prototype Documentation on an earlier/outdated plant air filter prototype

The biobroth bubbler (pictured above) is still in its experimental stage and we are looking into the potential harms of introducing these bacteria colonies in the home before posting further documentation. As a result this wiki focuses on the rhyzo-remediatior (plant air filter).

Remediation

Domestic chemical ecologies have both many toxicant sources and many toxicant sinks. Formaldehyde slowly and silently off gasses from engineered woods, carpets, and permanent press clothing. Some of these silent emissions waft out open windows. Others are absorbed by human bodies or the bodies of companion species. Others still are metabolized by decorative indoor plants and the microbes that inhabit their roots.

Since the 1990s we have been hearing promises of plant-based air filters. See, for example, this patent filed in 1993. Some critics (see this 2009 critical review) find fault in the claim that plants-on-their-own can clean indoor air and state that the average home would need 680 plants to adequately scrub the air of toxicants.

Increasing the Airflow

Research indicates that it is not so much that different plants remediate the air differently, but that different plants cultivate different rhizospheric conditions by their root exudates and root autolysis. Plants with quantitatively more bacteria dwelling in their rhizosphere won’t remediate as much atmospheric formaldehyde as plants with fewer overall microorganisms amongst their roots but more gram-negative rods. The formaldehyde removal capacity of plants and their bacterial companion species increases as exposures increase, and over time the bacteria appear to “acquire an increased taste” for exposed toxicants and up-regulate their metabolism. Phyto-bio remediation does not appear to be reaction-limited but diffusion-limited. The amount of air breezing through the roots and across the leaves of indoor plants is minimal. The Public Lab DIY Indoor Air Remediation Kit builds upon these observations that, at least in theory, increasing the airflow through the root system could increase the remediation capacity of plants by upwards of 200 fold. Plant air scrubbers would not require the intermittent replacement replacement of filters but would become more efficient overtime.

Plants manage decomposition underground by providing sugars to symbiotic and beneficial fungi and bacteria in order to manage root growth and decomposition. Plants spend more energy below ground than above ground, and seasonally grow and kill roots just like leaves. We're piggy backing on this process, using plant-managed colonies of beneficial bacteria to scrub formaldehyde, a naturally occurring product of decomposition.

While there is evidence on both sides to indicate either the efficacy or inefficacy of plants-on-their-own as indoor air filters, there have been no published studies on the efficacy of plant systems with increased airflow, like the prototype offered by Public Lab. A similar $230 system made by a for profit company claims that it can clean a 10'x14' room in an hour. Some interior-landscaping industries suggests that 1 regular plant on its own (which could be 200 times less efficient than the Public Lab prototype) per 100 sq feet. critics say its more like 45 regular plants per 100 feet (surprise! you live in a greenhouse now!) so that would be 680 plants in a 1500 sq foot house. But if our system lives up to the theory you could have just three planters to maintain clean air across your home or one in your bedroom or one in the office. In the first field test of our remediation tool we found a 40% reduction of formaldehyde in a two bedroom trailer with a single planter. In the second field test we found and an 84% decrease in a conventional two bedroom home. Pick Your Plant For our system you will need to add a plant -- we suggest Golden Pathos or Snake Plant (Sansevieria) because of their metabolic efficiency and hardiness. Here are a few other plants that have been show to reduce ambient formaldehyde loads (starred * items also metabolize toluene and xylene): • Dwarf date palm * • Boston fern * • Kimberly queen fern * • English ivy * • Lilyturf * • Spider plant * • Devil's ivy * • Peace lily * • Flamingo lily * • Chinese evergreen • Bamboo palm * • Broadleaf lady palm * • Variegated snake plant * • Heartleaf philodendron • Selloum philodendron • Elephant ear philodendron • Red-edged dracaena * Cornstalk dracaena Weeping fig * • Barberton daisy • Florist's chrysanthemum * • Rubber plant Current prototype See our first field test results here, yielding a 40% reduction of formaldehyde after a month. Our second field test, which more closely follows our design protocol, can be found here, and resulted in an 84% reduction in formaldehyde and a reduction in signs and symptoms of exposure after 12 weeks. Supplies: 1. You'll need to buy an aquarium pump (the$5.59 model 10 here) and reverse the diaphragm inside after removing the blue shell with a phillips head screwdriver. Here is a looping gif showing this simple hack (your pump will have one diaphragm in it and not two as is shown in the gif).

You can find this model at the above link or in most chain pet supply stores although it will be a few dollars more expensive in a brick and mortar store.

1. A few feet of flexible airline tubing ($1.99), available widely. 2. A non-draining flower pot. 3. A plant (see the above list). 4. A hydroculture growth medium such as Growstone and a 25 oz jar of activated carbon ($11-$20) from the fish aisle at the pet store). The growth medium will ship in units often much larger than what you will need for this project. Ask your local nursery or flowershop if you can buy as much as you need for your pot (should be less than$10).

For a 1500 sq. ft. home three plants might be best but one plant in one room that you spend a lot of time in or are particularly concerned about is a good way to start.

Instructions:

When you get the growth medium there is a little bit of work you have to do to get the plant ready. You can imagine how jarring it must be for a plant to suddenly have air rapidly flowing through its roots. So these first few steps ensure that your plant will be healthy so that your air can be healthy. It would be great if you could take pictures of the steps and we can help you along the way. Once you get the plant happy, all you need to do is periodically water it, and its metabolic capacity will actually increase with time (unlike mechanical filters that decrease with time). The aquaculture set up also decreases the amount of watering you have to do.

1. Soak the activated carbon and the ceramic media in water to saturate. (Can be combined in a large bucket.) Soak a minimum of one hour. Both the carbon and the media absorb and hold moisture. The activated carbon will make noise when initially submerged in water. This is completely normal. After soaking the media, rinse with water until water runs clear. (The inner pot is often used as a strainer for this process.) Use a doubled cheesecloth or an old T-shirt to strain the carbon. Once the water runs clear, if you have not already done so, combined both the ceramic and the carbon in clean water and mix thoroughly. Mixing the media is easier to do when submerged in water.

Photo by Tim Choy

1. Remove the plant from its pot and gently remove as much soil from the roots as possible. Soak the plant’s roots in room temperature water for a short time. This will help loosen the remaining soil and make it easier to remove. After soaking, gently agitate the roots in the water to help remove soil. Gently massaging the root ball in the water will remove a significant amount of the soil. Using a fresh pail of room temperature water, a shower-head or faucet, clean the roots of all remaining soil, being very careful not to damage the root system. Repeat this process until the roots are entirely free of soil. This is very important for the future integrity of the plant!

Photo by Jerry Zee

1. Once your root system is entirely free of soil, cut back roughly a quarter of the root system with a pair of sharp scissors. This will stimulate growth of new “water roots.” Note: keep root system moist while transplanting; do not allow roots to dry out. Now that your plants roots are entirely free of soil and your media is free of dust, you are ready to begin potting.

2. Pot the plant as you normally would just within the growth medium you have prepared. Be sure to snake the aquarium tubing over the lip of the container, with the end terminating high in the root ball. You will want to never have the water level in your planter high enough to be sucked into the tubing and destroy or contaminate the pump.

Allow your plant to acclimate itself to its new environment for at least two weeks before turning on the pump.

Instructions are adapted from the plans of former NASA scientist, Dr. Bill Wolverton.

First prototype

This was developed in a workshop at App State University in the fall of 2014. The setup adapts an existing pot with holes in it (this patching technique is not advised as it may introduce VOCs into the system) and also uses a flowmeter to assess the flow rate of the system, which is not necessary for daily use.

Start By sealing all but one hole in the bottom of the flower pot. (This open hole will be used to insert the aquarium tube) Make sure that the holes are tightly sealed to avoid water leakage.

Insert one section of the tubing in the hole (roughly 5" into the pot. avoid crimping the tube), and seal the hole around the tube with the epoxy.

Fill the pot with 3 inches of growth medium ensuring that the tube is above the rocks.

Attach the airstone intake to the tube inside the pot and cover with more rocks. Mark on the outside of the pot where the bottom of the intake is. This will act as a fill line when watering. If water is above this line, it will get sucked into the intake.

Cover the intake with another 1 in of rocks or growth medium of choice.

Next, place the plant on top of the rocks and secure more rocks or growth medium around the roots of the plant. Secure the plant around the roots. (Leave an inch in between the root systems and the air intake system to avoid suffocating the filter without losing air flow within the plant system.)

Next, attach a valve connector to the outside part of the tube. This is the end coming out of the pot. You can run any necessary length of tubing from this connector to the intake of the flow meter (the intake and outtake of the flow meter should be equipped with the conversion valves mentioned in materials.)

Next attach the tubing from the outtake to the vacuum pump. See below on how to convert the aquarium pump into a vacuum pump.

Reversing Pump and setting up flow meter with valves.

The first step is to reverse the diaphragm in the pump to convert it to a vacuum. Open up the four Philips head screws on the belly of the pump. Lift up, rotate 180 degrees, and then replace the diaphragm as indicated in the below GIF.

https://i.publiclab.org/system/images/photos/000/007/694/original/output_ra01xV.gif

Attach a small segment of airline tubing from the pump to the top of the flow meter (which you have equipped with the conversion valves at both openings) and then attach another small section of tubing to the bottom of the flow meter. You will have three points where you can adjust the flow rate (the meter, and both conversion valves). Open all valves to their most open--while still sealed--position. Once you've started the pump, use the large valve on the flow meter to regulate the flow rate down to .3 liters per minute (or 300 ml per minute).

Notes (to be cleaned up later): Gently separate (we're using mother-in-law's tongue) plants, remove dirt from plants We then filled the pot with recycled glass 3" up from the top, placed our filter in it, then filled another inch of rocks, placed the plant, and then filled the pot with rocks.

test air flow w/ flow meter each configuration before placing in the pot- 0.65 liters per minute (discard-a-stone coarse bubble), 0.7 liters per minute (aqua mist 1 cm), 0.65 liters per minute (aqua mist 5 cm x 14 cm)

Tube length has a negligible effect on air flow- we tested both short (4") and long (6') and saw little difference between the two. Is there a limit for how much air flow is too much for the plant? (temperature difference, etc.)

epoxy to fill in the holes of the plastic pot that we're using (BE SURE TO THOROUGHLY SEAL HOLES), seran wrap should work as well; use one of the existing holes to run the tube through the pot. It's useful to find a pot that has sufficient gaps to run the tubing underneath without crimping. Leave enough room with the tube to have at least a few inches running outside of the pot, Make sure to have enough tubing to allow the filter used to sit above the water level.

Leave an inch in between the root systems and the air intake system to avoid suffocating the filter without losing air flow within the plant system. With this setup, we're getting 0.95 liters per minute of airflow.

to reverse the motor remove diaphragm from the pump and reverse it- vacuum bingo! (GIF LOCATION)

http://www.wolvertonenvironmental.com/airFAQ.htm


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# Fill in the blank: A function f : X → Y is said to be ______, if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f (x) = y.

## 1 Answer

Answer: Onto or Surjective.

# Fill in the blank: A function f : X → Y is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f (x) = y.

answered Mar 12, 2013
True or False: A Function can be both one-one and onto.

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer


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# Convergence of a Cauchy sequence of matrices

I have a Cauchy sequence of matrices $C_i \in R^{p \times q}$, i.e. $\lim_{n\rightarrow \infty} \| C_{n+1} - C_{n} \| = 0$ for any norm (I just need the property that $\|C_1-C_2\|>\delta \Rightarrow C_1 \neq C_2$). I also know $\|C\|_{1,1} \leq t$ for a fixed $t$ (where $\|C\|_{1,1} = \sum_{i,j} |c_{ij}|$).

Can I conclude that the sequence $C_i$ also converges? (i.e., is the corresponding metric space complete?)

• What you describe (the limit of $\| C_{n+1}-C_n\|$ being $0$) does not mean that $C_n$ is Cauchy, and it doesn't imply convergence of the sequence. It is true however that the space of $p\times q$ real matrices is complete. – Olivier Bégassat Jul 30 '14 at 17:06

If a sequence in this space is Cauchy, then it converges. That is, $\Bbb R^{p \times q}$ under $\|\cdot\|_{1,1}$ (which is isometric to $\Bbb R^{pq}$ under $\|\cdot\|_1$) is indeed a complete metric space.
That being said, the condition you provided is insufficient to guarantee that the sequence is Cauchy. As a counterexample, consider the sequence in $\Bbb R^{2 \times 1}$ given by $C_n = (\cos \theta_n,\sin\theta_n)$ where $$\theta_n = \sum_{i=1}^n \frac 1i$$ Noting that $\|C_n\| < \sqrt{2}$ for each $n$.
• You could a sequence whose sequence of partial sums oscillates between $0$ and $1$, for instance, $1,-1/2,-1/2,1/4,1/4,1/4,1/4,-1/8,-1/8\dots$ – Olivier Bégassat Jul 30 '14 at 17:20
• @Milad it's equivalent to consider sequences of the form $\sum_{i=1}^n D_i$ where $D_i = C_{i+1} - C_i$. So, what you should really be looking into is the convergence of series. – Omnomnomnom Jul 30 '14 at 19:17


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# equation for an hexagon with rounded corners

I know a simple equation for a squircle, $$x^4+y^4=a^4$$

What would be the equation for an hexagon with rounded corners?

In the equation, how can I take in account for the angle of rotation of the hexagon about its center?

The following equation gives a rounded hexagon where $$\theta$$ can be changed in order to give any rotation of the hexagon about its centre and $$r$$ can be changed to affect the roundness of the hexagon itself. I found that $$r=15$$ works well. The value $$10$$ at the end of the equation can be increased to increase the size of the hexagon as well. $$\sum_{n=1}^6 \Big|x\cos{\Big(\frac{\pi n}{3}+\theta\Big)}+y\sin{\Big(\frac{\pi n}{3}+\theta\Big)}-\frac{1}{6}\Big|^r=10$$
polar $$r = 5 + \frac{1}{5} \cos (6 \theta)$$ looks very good. You can solve as rectangular using $$x = r \cos \theta$$ and $$y = r \sin \theta$$


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## Precalculus (6th Edition) Blitzer

$\{\frac{\pi}{4},\pi,\frac{5\pi}{4} \}$
Step 1. The equation gives two solutions: $tan(x)=1$ and $cos(x)=-1$ Step 2. For $tan(x)=1$, we have the reference angle as $x=\frac{\pi}{4}$ and we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{4},\frac{5\pi}{4}$ Step 3. For $cos(x)=-1$, we can find all x-values in $[0,2\pi)$ as $x=\pi$ Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{4},\pi,\frac{5\pi}{4} \}$


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Covalency and Oxidation Number in Covalent Compounds

1. Feb 14, 2013

hms.tech

1. The problem statement, all variables and given/known data

$NCl_{3}$

What is the Co-valency / Oxidation number of Nitrogen in this Covalent compound ?

2. Relevant equations

I think the table of electronegativity might be useful

3. The attempt at a solution

Nitrogen must be "-3" since it is more electronegative than Chlorine.

The correct answer is +3. How is that possible ?

Clearly Nitrogen is more electronegative than chlorine.

2. Feb 14, 2013

Saitama

Chlorine is more electronegative than nitrogen. :)

3. Feb 15, 2013

hms.tech

Nitrogen forms hydrogen bonds in its compounds with Hydrogen attached .

Chlorine rarely forms any hydrogen bonds in any of its compounds .

Here :

4. Feb 15, 2013

Saitama

Allen Scale?

Pauling Scale is the one you need to use.

5. Feb 15, 2013

hms.tech

hmm...doesn't Pauling scale gives us wrong values since according to pauling scale Chlorine being more Electronegative should form Hydrogen bonds .

6. Feb 15, 2013

Staff: Mentor

As if it all mattered...

First: oxidation numbers don't reflect any real property of the element, they are used just for electron accounting.

Second: using electronegativity to predict anything is a quite fuzzy concept, more of a general indication than a hard rule, especially when the differences in electronegativity are small.

Taking both things into account - don't be surprised you get inconsistent predictions trying to apply rule of thumb to border cases.


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# Exam question: Normals

While solving exam questions, I came across this problem:

Let $M = \{(x, y, z) \in \mathbb{R^3}; (x - 5)^5 + y^6 + z^{2010} = 1\}$. Show that for every unit vector, $v \in \mathbb{R^3}$, exists a single vertex $p \in M$ such that $N(p) = v$, where $N(p)$ is the outer surface normal to $M$ at $p$.

I don't know where to start. Ideas?

-
Are you sure you copied the question correctly? This surface intersects every line parallel to the $x$ axis. I suspect the $x-5$ should have an even exponent. – Robert Israel Jul 17 '12 at 18:00

First, note that if $z=0$, then the normal will also have a zero $z$ component, so we can take $z=0$ and look for issues there. Plotting the contour of $(x-5)^5+y^6 = 1$ gives the picture below:
Pick a normal $v$ to the surface around $(2,2.5)$. Then note that in the vicinity of $(6,1)$, it is possible to find a normal that matches $v$.


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## Texas Clean Energy Project signs long-term CO2 offtake agreement with Whiting Petroleum for enhanced oil recovery; 90% CO2 capture from IGCC coal polygen plant

##### 22 July 2011
 The TCEP would integrate coal gasification, combined-cycle power generation, CO2 capture, and urea production. CO2 capture and shipment via pipeline shown at top. Click to enlarge.

Summit Power Group, LLC and Blue Strategies, LLC have signed a 15-year CO2 offtake agreement with Whiting Petroleum Corporation for some of the CO2 that will be captured by Summit’s Texas Clean Energy Project (TCEP), a coal gasification with carbon capture polygen project being developed in Penwell, Texas, outside Odessa.

TCEP will be an integrated gasification combined cycle (IGCC) 400 MW power/poly-gen plant that will capture 90% of the carbon dioxide, 99% of the sulfur, more than 95% of the mercury, and eliminate more than 90% of the nitrogen oxides produced by the process. Its high carbon-capture rate reduces the total amount of CO2 emitted to the atmosphere to less than 10% of the emissions for power from a conventional coal plant, and less than 25% of those for power from a highly efficient, natural gas-fired plant, according to Summit.

Under the new agreement, Whiting will purchase 80 million cubic feet (80,000 Mcf/d) of compressed CO2 per day from TCEP—representing approximately 60% of TCEP’s total volume of captured CO2—during the first five years of TCEP’s operation, with gradually declining amounts and an option to extend the purchases thereafter. In the Permian Basin, approximately one additional barrel of oil can be recovered for each 6,000 cubic feet (6 Mcf) of compressed CO2 injected into the oil field. TCEP will begin delivering CO2 to Whiting when the plant commences operations in late 2014 or early 2015; construction is scheduled to begin at the end of this year.

This is another important milestone for the Texas Clean Energy Project, coming on the heels of last month’s power purchase announcement with CPS Energy in San Antonio. We now have sales commitments in place for all three of TCEP’s main commercial products—electric power, urea for fertilizer, and CO2 for enhanced oil recovery—and that is obviously key to getting this project underway.

—Donald Hodel, Chairman of Summit Power

CO2 for EOR operations in the Permian Basin and elsewhere have generally been carried out with CO2 that is geological in origin—i.e., formed naturally in underground reservoirs in New Mexico and Colorado and brought to the surface by wells such as those used to produce natural gas. More than 3,000 miles of CO2-dedicated pipeline currently transports and distributes geologic CO2 throughout the Permian Basin.

Whiting will be the first in the Permian to purchase CO2 from a power project that will be produced through the coal-gasification process. TCEP received a $450 million award in 2010 from the US Department of Energy’s Clean Coal Power Initiative (CCPI), the Department’s effort to create a new generation of energy processes that sharply reduce emissions from coal-fired power plants and reduce America’s dependence on imported energy resources. (Earlier post.) In oil fields such as Whiting’s, the injected CO2 mixes with the oil that is left behind in the primary oil-well production and the secondary water-injection stage. Approximately 40% of the initially injected CO2 remains trapped underground. The remainder comes to the surface with the oil, but is then recaptured, recompressed, and re-injected. Ultimately, some 99% of the injected CO2 can be permanently stored (i.e. geologically sequestered) deep underground many thousands of feet below the water table, with no leakage to the atmosphere. We are extremely pleased to be the first oil producer in West Texas to start using manmade CO2 from coal for enhanced oil recovery. This shift from geologic to manmade CO2 in the oil fields is a significant step forward for both the power industry and the oil industry. —Whiting CEO James Volker The TCEP The TCEP integrates coal gasification, combined-cycle power generation, CO2 capture, and urea production. Siemens is the primary equipment provider for TCEP’s gasifiers, power island and controls, including a “twin pack” of SFG-500 gasifiers, a SGT6-5000F combustion turbine, and an advanced SPPA-T3000 control system. The project’s front end engineering design (FEED) study was launched in June 2010 by Fluor, Siemens and Selas Fluid Processing Corporation, a Linde Group subsidiary. The project is supported by H.B. 469, Texas legislation enacted in 2009 to promote carbon capture power projects with low emissions. Coal gasification, syngas processing and CO2 capture. Pulverized coal feedstock will be introduced into the two Siemens gasifiers along with limited amounts of nearly pure O2 gas and converted into syngas comprising H2 and CO, varying amounts of CO2, nitrogen (N2), sulfur species, methane, volatilized metals, and PM. The syngas will be cooled and cleaned of PM. Next, the syngas flows through a water-gas shift reactor, in which steam is injected in the syngas over a catalyst bed, initiating a reaction where the CO in the syngas would be converted to CO2 and the steam would be converted to additional H2 in the syngas stream. This provides a syngas stream that is concentrated in both CO2 and H2. Subsequently, the syngas would pass through a mercury removal system and then an acid gas removal system where first the sulfur species would be removed, then the CO2, creating a clean, H2-rich concentration syngas upon exiting the acid gas removal unit. Captured CO2 will be further cleaned and compressed, and then transported by pipeline to an existing regional CO2 pipeline or, potentially, to a nearby EOR field. A portion of the captured CO2 will also be used to produce urea. The H2-rich syngas stream will be split, with part used to produce electricity via the turbine and the other part be used to produce urea for fertilizer. Argon and H2SO4 are by-products of the gasification process and would be made available for commercial sale. Inert slag, another by-product of the gasification process, would be sold for manufacturing and construction uses or disposed of off-site. The H2-rich, low-CO2 syngas will be combusted in a turbine generator to produce electricity. Combustion of the H2-rich fuel gas will produce water vapor and a low-CO2 exhaust gas with significantly lower CO2 emissions than would occur if the coal itself, or the raw syngas, had been combusted. The exhaust gas would be ducted through an HRSG (heat recovery steam generator), which would generate high-temperature, high-pressure steam. This steam would be piped into a steam turbine-generator, which would generate additional electricity. This integration of the combustion turbine-generator, HRSG, and steam turbine-generator is known as a combined-cycle power plant. The combined power generation from the combustion turbine-generator and the steam turbine generator would be approximately 400 MW (gross) with 213 MW sent to the grid, on average, and the remainder being used to run the plant’s equipment. The electricity sold would be transmitted to the regional electrical grid by a high voltage transmission line system. Natural gas would be used to start up the polygen plant and as a backup fuel (natural gas would also be used during operations to heat drying gases, supply an auxiliary boiler, and provide burner pilot flames such as for flares). With two Siemens gasifiers, the TCEP will produce more syngas than can be used for electricity production. The additional syngas produced will be converted to NH3 using the Haber process. In that process, the H2 in the syngas is reacted with N2 from the air separation unit, forming NH3. Downstream, the NH3 is reacted with a portion of the CO2 from a syngas cleanup system, thereby forming urea in a Bosch-Meiser process. The urea is produced as a granular product common in the fertilizer industry. TCEP will be located at one of the former FutureGen finalist sites—the 600-acre Penwell site—situated fifteen miles west of Odessa, Texas. An environmental impact statement (EIS) is currently being prepared for the US Department of Energy (DOE). TCEP received its air quality permit from the Texas Commission on Environmental Quality (TCEQ) on December 28, 2010—a major milestone that is the key state governmental approval that is required to move the project forward to be privately financed and built. The permit was issued administratively since the draft permit received no opposition or requests for a hearing. TCEP is currently scheduled to achieve financial closing and commence construction in Fall 2011. Commercial operation is scheduled for late 2014. The project will begin sequestering carbon during start-up and testing in 2014. Summit, a power developer based in Seattle, WA, and Blue Strategies, a Houston-based developer of physical CO2 projects, partnered in October 2009 to market TCEP’s 2.5 million tons per year of CO2 to oil producers in the West Texas Permian Basin. This is the first of several CO2 off-take agreements with TCEP, and we are pleased to have negotiated this historic first one with a company that has a long and successful history of doing enhanced oil recovery in the Permian. —Blue Strategies Executive Vice President Russell Martin Whiting Petroleum Corporation is an independent oil and gas company that acquires, develops and explores for crude oil, natural gas and natural gas liquids primarily in the Permian Basin, Rocky Mountains, Mid-Continent, Gulf Coast and Michigan regions of the United States. The company’s largest projects are in the Bakken and Three Forks plays in North Dakota and its Enhanced Oil Recovery (EOR) projects, which have historically used naturally occurring geological CO2 for injection in its EOR projects in Oklahoma and Texas. Summit Power Group has developed large, low-carbon energy projects with more than 7,000 megawatts of electric power plants in operation, representing more$7 billion in investment, and more than 2,000 megawatts in development or under construction.

Blue Strategies is a leading developer of CO2 projects involving the capture, compression and transportation of CO2. In addition, Blue Strategies works very closely with private and governmental agencies to develop procedures around assuring injected CO2 remains sequestered in the reservoir and meets the requirements for carbon offset registries. The management team has more than 100 years of experience in managing physical CO2.

Resources

So if they add only a small percentage (>10%) of biomass, this would be truely carbon-negative electricity. even better than wind or solar.

Possibilities have been there for a long time.

The good news is that the market value of CO2 gasses increases with increasing oil prices as it can be used to squire even more oil out of aging oil fields. However, I am skeptical that it is economical to source the CO2 from coal power plants. It requires very expensive and complicated equipment to separate nitrogen from the CO2 in the exhaust gases of a coal plant. Why not do it with an ethanol refinery instead. As far as I understand the fermentation process in an ethanol plant produces pure CO2 gasses without nitrogen or any other gases in it. No equipment is needed apart from a pump and a CO2 pipeline to the nearest oil field.

In all likelihood Poet energy will succeeds in making cellulosic ethanol economic no later than 2015. This will increase the potential to locate ethanol refineries on top of old oil fields as cellulosic biomass can be grown almost anywhere as long as you have water. You don’t need fertile land to grow grasses for ethanol production. The point is off cause to minimize the cost of pumping CO2 to these oil fields.

400 MW gross with only 213 MW net is one heck of a lot of plant overhead. The syngas split between generation and chemical production may account for much of that, and CO2 compression some more, but that is amazingly high. I've been touting concepts like this for some time, but if the output fraction doesn't improve greatly if the chemical side is removed then it just isn't viable due to low efficiency. (Wabash River hits 40%, w/o CO2 capture.)

I am skeptical that it is economical to source the CO2 from coal power plants. It requires very expensive and complicated equipment to separate nitrogen from the CO2 in the exhaust gases of a coal plant.
Henrik, the CO2 in this plant is captured from the acid-gas scrubbers handling the output of oxygen-blown gasifiers. There is little nitrogen involved.
Why not do it with an ethanol refinery instead. As far as I understand the fermentation process in an ethanol plant produces pure CO2 gasses without nitrogen or any other gases in it. No equipment is needed apart from a pump and a CO2 pipeline to the nearest oil field.
Not a bad idea. Maybe you can find a suitable oil field and persuade a distillery to locate nearby.

The project is supported by H.B. 469, Texas legislation enacted in 2009 to promote carbon capture power projects with low emissions.

Texas is certainly not interested in CO2 storage. Sounds very big and noisy. Why not clean up methane and burn it? In Texas just build your plant near the field - use the carbon stream for EOR.

The hot exhaust of the gas plant would probably enhance oil recovery... and the gas fields are air tight, over geologic times so there would be some CO2 mitigation after all.

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Hello All, This is my first post. I have been writing software for embedded systems for 10 years. I have C++ knowledge, but very rusty. I am wondering if anyone has experience with the GameInstitue classes, the C++ class in particular. I am wondering if it is worth the $70. I have the Ivor Horton teach yourself VC++, but have been having a hard time doing that instead of Day of Defeat. :-) I have decided to try my hand at writing a game. Any feedback on GameInstitute would be helpful. Thanks, rallen911 (That''s Porsche 911, not 9/11) #### Share this post ##### Link to post ##### Share on other sites Advertisement It seems like a lot of the stuff in those courses can be found in books written by the faculty that teach them. Or you can just google up each of the topics and see what tutorials you find for free. GameInstitute seems kinda pricey to me. You can get pretty much the same interaction here at the GameDev forums. #### Share this post ##### Link to post ##### Share on other sites I have taken the c++ tutorial on gameinstitute and i am realy happy that i did. #### Share this post ##### Link to post ##### Share on other sites I would agree that I can probably find the information for free. My main reason for considering the Game Institue is having more of a stake in the process. If I fork out$70 to take the class, I am more likely to ''attend'' the lectures, read the material, do the exercises.

My main problem is a character flaw in that I have trouble motivating myself to do that sort of stuff. Even though it is fun, I have already been writing software for a good part of the day.

Thanks for the feedback.

rallen911

If you need to be in the classroom environment then $70 bucks for a C++ class doesn''t seem that bad, given what colleges are charging per credit hour these days. But seriously if you just have to be in a classroom, then I will be your teacher. Once you send me three easy payments of$9.99 class will begin.