text:
### Mutual inductance - class-XII

 Description: mutual inductance Number of Questions: 94 Created by: Jayanti Mahajan Tags: physics electromagnetic induction and alternating currents electromagnetic induction
Attempted 0/94 Correct 0 Score 0

The coefficient of mutual inductance between two coils depends on

1. medium between the coils

2. separation between the two coils

3. orientation of the two coils

4. all of the above

Correct Option: D
Explanation:

The flux linked with two coils will depend upon the angle between the two coils. If their planes are parallel, then magnetic flux from one would completely  pass through the other. If the planes are perpendicular, no flux due to any of the coils would flow through the other.

The size of the two coils may be different which will affect the number of lines crossing the coil. The medium, if magnetic, will concentrate the field lines. Thus, all parameters would affect the inductance between them.

Two coils of self inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

1. 10 mH

2. 6 mH

3. 4 mH

4. 16 mH

Correct Option: C
Explanation:

Given,

$L _1=2mH$
$L _2=8mH$
The mutual inductance between coil is
$M=\sqrt{L _1L _2}$
$M=\sqrt{2\times 8}=\sqrt{16}mH$
$M=4mH$
The correct option is C.

A circular copper disc 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of 1 Wb $m^2$ is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim ?

1. 0.023 V

2. 0.23 V

3. 23 V

4. 230 V

Correct Option: B
Explanation:

Here,

$l = r = 5\, cm = 5 \times 10^{-2} m,$
B = 1 Wb $m^{-2}$

$\omega = 2 \pi \left( \dfrac{1800}{60} \right) \, rad \, s^{-1} = 60 \pi \, rad \, s^{-1},$

$\epsilon \, = \, \dfrac{1}{2} Bl^2 \omega \, =\, \dfrac{1}{2} \times 1 \times (5 \times 10^{-2})^2 \times 60 \pi = 0.23 V$

Mutual inductance of two coils can be increased by

1. decreasing the number of turns in the coils

2. increasing the number of turns in the coils

3. winding the coils on wooden cores

4. none of these.

Correct Option: B
Explanation:

As M  = $\dfrac{\mu _0N _1N _2A}{l}$
i.e M can be increased by increasing the numbers of turns in the coils.

If the self inductance of 500 turns coil is 125 mH, then the self inductance of the similar coil of 800 mH

1. 48.8 mH

2. 200 mH

3. 290 mH

4. 320 mH

Correct Option: D
Explanation:

$L=\mu _o \mu _r N^2Al$
For similar coil, $A, l$ will be same
So,   $\, \dfrac{L _1}{L _2} = \dfrac{N _1^2}{N _2^2}$

$L _{800}= \dfrac{N _{800} ^2}{N _{500}^2}\times L _{500}= \, \dfrac{125}{(500)^2} \, \times \, (800)^2 \, = \, 320 \, mH$

The mutual inductance $M _{12}$ of a coil 1 with respect to coil 2

1. increases when they are brought nearer

2. depends on the current passing through the coils.

3. increases when one of them is rotated about an axis.

4. both (a) and (b) are correct

Correct Option: A
Explanation:

Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.

If the two coils $1$ and $2$ are present with mutual inductance $M _1$ and $M _2$. Then the mutual inductance of the coil 1 due to 2 increases when they are bought near since, mutual inductance is  proportional to the flux passed through the coil.

The mutual induction of $M _{12}$ is same as $M _{21}$

Match the following:

Quantity Formula
1) Magnetic flux linked with a coil a) $\displaystyle -N\frac { d\phi }{ dt }$
2) Induced emf b) $\displaystyle { \mu } _{ r }{ \mu } _{ 0 }{ n } _{ 1 }{ n } _{ 2 }{ \pi r } _{ 1 }^{ 2 }l$
3) Force on a charged particle moving in a electric and magnetic field c) $\displaystyle BA\cos { \theta }$
4) Mutual inductance of a solenoid d) $\displaystyle q\left( \overline { E } +\overline { v } \times \overline { B } \right)$
1. 1-c, 2-d, 3-b, 4-a

2. 1-c, 2-a, 3-d, 4-b

3. 1-b, 2-a, 3-c, 4-d

4. 1-a, 2-b, 3-d, 4-c

Correct Option: B
Explanation:

1) Magnetic flux through any area is the scalar product of its area vector with the magnetic field vector. Thus for a coil, it is $\vec{B}.\vec{A}=BAcos\theta$

2) Emf induced in a coil due to changing flux through it is given by Faraday's Law,
$Emf = -N\dfrac{d\phi}{dt}$
3) Force on a charged particle due to electric field = $q\vec{E}$
Force on a moving charged particle due to magnetic field = $q(\vec{v}\times \vec{B})$
Thus, force on a moving charged particle in an electric and magnetic field = $q(\vec{E}+\vec{v}\times \vec{B})$
4) Mutual inductance of a solenoid is found out to be : $\mu _r\mu _0n _1n _2\pi r _1^2l$

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in secondary in the step-up transformer is $1:10$. If the power to the consumer has to be supplied at $200\ V$, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is:

1. $200:1$

2. $150:1$

3. $100:1$

4. $50:1$

Correct Option: A

An inductor of inductance $100\ mH$ is connected in series with a resistance, a variable capacitance and an AC source of frequency $2.0\ kHz$; The value of the capacitance so that maximum current may be drawn into the circuit.

1. 50 nF

2. 60 nF

3. 63 nF

4. 79 nF

Correct Option: C
Explanation:

$\begin{array}{l}{X _L} = Lw = {10^{ - 1}} \times 2\pi \times 2 \times {10^3}\{X _L} = 4\pi \times {10^2}\Z = \sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} \i = \dfrac{V}{Z} = \dfrac{V}{{\sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} }}\for,{i _{\max }}\{X _L} = {X _C}\\therefore {X _C} = Lw = \dfrac{1}{{Cw}}\C = \dfrac{1}{{{w^2}L}} = \dfrac{1}{{{{10}^{ - 1}} \times 4{\pi ^2} \times 4 \times {{10}^6}}}\ = \dfrac{{{{10}^{ - 5}}}}{{16{\pi ^2}}} = 63nF\end{array}$

$5 \mathrm { mV }$ is induced in a coil, when current in another nearby coil changes by $5 \mathrm { A }$ in $0.1$sec. The mutual inductance between the two coils will be

1. $0.1 \mathrm { H }$

2. $0.2 \mathrm { H }$

3. $0.1 \mathrm { mH }$

4. $0.2 \mathrm { mH }$

Correct Option: A

In mutual induction
A: when current in one coil increases, induced current in neighbouring coil flows in the opposite direction
B: When current in one coil decreases, induced current in neighbouring coil flows in the opposite direction

1. A is true, B is false

2. A and B are false

3. A and B are true

4. A is false, B is true

Correct Option: A

In case of all flux from the current  in coil 1 links with coil 2, the coefficient of coupling will be

1. 2.0

2. 1.0

3. 0.5

4. zero

Correct Option: B
Explanation:

If all flux from the current in coil 1 links with coil 2, then that is referred as an ideal transformer. For an ideal transformer, the coefficient coupling is 1.

The induction coil works on the principle of.

1. Self-induction

2. Mutual induction

3. Ampere's rule

4. Fleming's right hand rule

Correct Option: B
Explanation:

Induction coil works on the principle of mutual induction that an emf or current is induced in the second coil if the magnetic flux due to first coil linked with the second coil changes.

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of $10^{-3}$ Wb to link with A and a flux per turn of $0.8 \times 10^{-3}$ Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is :

1. 5/4

2. 1/1.6

3. 1.6

4. 1

Correct Option: B
Explanation:

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10−3 Wb to link with A and a flux per turn of 0.8×10−3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is $\dfrac{L _1}{L _2}=\dfrac{200*10^{-3}}{400*0.8*10^{-3}}=1/1.6$

Two concentric coils each of radius equal to $2\pi\ cm$ are placed at right angles to each other. $3$ Ampere and $4$ ampere are the currents flowing in each coil respectively. The magnetic induction in $Weber/m^{2}$ at the centre of the coils will be ($\mu _{0}=4\pi \times 10^{-7}\ Wb/A-m$)

1. $12\times 10^{-5}$

2. $10^{-5}$

3. $5\times 10^{-5}$

4. $7\times 10^{-5}$

Correct Option: C
Explanation:
Magnetic Induction at centre of coil is
$B=\dfrac {\mu _0}{2r} \sqrt {I _1^2 +I _2^2}\quad I _1=3\ A$
$I _2=4\ A$
$=\dfrac {4\pi\times 10^{-7}}{2\times \dfrac {2\pi}{100}}\times \sqrt {3^2 +4^2}$
$=\dfrac {4\pi \times 10^{-5}\times 5}{2\times 2\pi}$
$=5\times 10^{-5} \dfrac {wb}{m^2}$

A 60 volt - 10 watt bulb is operated at 100 volt - 60 Hz a.c. The inductance required is?

1. 2.56 H

2. 0.32 H

3. 0.64 H

4. 1.28 H

Correct Option: D

Two coaxial coils are very close to each other and their mutual inductance is $5mH$. If a current $50sin{500t}$ is passed in one of the coils then the peak value of induced emf in the secondary coil will be

1. $5000V$

2. $500V$

3. $150V$

4. $125V$

Correct Option: D
Explanation:

According to principle of mutual inductance, flux induced in coil  is equa to the current flowing in coil 1

$\phi _{2}= Mi _{1}$
$\dfrac{d\phi _{2}}{dt}=M\dfrac{di _{1}}{dt}$ = EMF
$\therefore EMF = 5\times 10^{-3}\dfrac{d}{dt}50 sin 500t$
$\therefore EMF = 125 cos500t$
So, maximum vaue of EMF would be 125 V

The coefficient of self induction of two inductor coils are $20mH$ and $40mH$ respectively. If the coils are connected in series so as to support each other and the resultant inductance is $80mH$ then the value of mutual inductance between the coils will be

1. $5mH$

2. $10mH$

3. $20mH$

4. $40mH$

Correct Option: B
Explanation:

$L _{total} = L _1 + L _2 + 2M$

$80 mH = 20 mH + 40 mH + 2M$

$\therefore M = 10 mH$

For solenoid keeping the turn density constant its length makes halved and its cross section radius is doubled then the inductance of the solenoid increased by :

1. $200\%$

2. $100\%$

3. $800\%$

4. $700\%$

Correct Option: B

A rectangular loop of sides 'a' and 'b' is placed in the XY plane. A very long wire is also placed in xy plane such that side of length 'a' of the loop is parallel to the wire. The distance between the wire and the nearest edge of the loop is 'd'. The mutual inductance of this system is proportional to?

1. a

2. b

3. $1/d$

4. Current in wire

Correct Option: A

A circular loop of radius $r$ is placed at the centre of current carrying conducting square loop of side $a$. If both loops are coplanar and $a >> r$, then the mutual inductance between the loops will be:

1. $\dfrac{\mu _0r^2}{2\sqrt{2}(a)}$

2. $\dfrac{\mu _0r^2}{4a}$

3. $\dfrac{2\sqrt{2}\mu _0r^2}{\pi a}$

4. $\dfrac{\mu _0r^2}{4\sqrt{2}a}$

Correct Option: C
Explanation:

Both loops are coplanar.

Magnetic field at the center of outer square current carrying loop is
${ B } _{ 1 }=\dfrac { 2\sqrt { 2 } { \mu } _{ 0 }I }{ \pi a }$
where $a$= length of side of square loop and
$r$= radius of the circular loop.
Given $a>>r$,
The magnetic field through entire inner coil is ${ B } _{ 1 }$
Magnetic flux through inner coil, ${ \phi } _{ 21 }={ B } _{ 1 }{ A } _{ 2 }$
=$\dfrac { 2\sqrt { 2 } { \mu } _{ 0 }I }{ \pi } \dfrac { { r }^{ 2 } }{ a }$------ (1)
Mutual induction, M= $\dfrac { \phi }{ { I } _{ 1 } }$

From (1), M= $\dfrac { 2\sqrt { 2 } { \mu } _{ 0 } }{ \pi } \dfrac { { r }^{ 2 } }{ a }$

Hence, $M\alpha \dfrac { { r }^{ 2 } }{ a }$

A $50\ Hz$ $AC$ current of crest value $1\ A$ flows, through the primary of transformer. If the mutual inductance between the primary and secondary be $0.5\ H$, the crest voltage induced  in the secondary is

1. 75 V

2. 150 V

3. 100 V

4. 300V

Correct Option: D

Which of the following statement is correct?

1. when the magnetic flux linked with conducting loop is zero then emf induced is always zero

2. when the emf induced in conducting loop is zero, then the magnetic flux linked with the loop must be zero

3. transformer works on mutual induction

4. all of these

Correct Option: A,C
Explanation:

Statement is.

A) When the magnetic flux linked with conducting loop is zero then emf induced is always zero.
$emf=\dfrac{d\phi}{dt}$
If $\phi=0$, $emf=\dfrac{d0}{dt}=0$
B) when the emf induced in conducting loop is zero, then the magnetic flux linked with the loop must be zero.
$emf=\dfrac{d\phi}{dt}=0$
$d\phi=0$
$\phi=constant$ magnetic flux is constant.
This is the wrong statement
C) The transformer works on mutual induction.
The correct statement is (A) and (C).

An electron originates at a point $A$ lying on the axis of a straight solenoid and moves with velocity $v$ at an angle $\alpha$ to the axis. The magnetic induction of the field is equal to $BA$ screen is oriented at right angles to the axis and is located at a distance $1$ from the point $a$. Find the distance from the axis to the point on the screen into which the electron strikes.

1. $d = 5r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = 2\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

2. $d = 2r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

3. $d = 3r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = 3\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

4. $d = 4r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

Correct Option: B

When current breaks in primary coil current reaches to zero in  second. Emf induced in the secondary coil is 20,000V and mutual inductance between the coils is 5H. The maximum current is the primary before the break is

1. 0.2 amp

2. 0.4 amp

3. 4 amp

4. 2 amp

Correct Option: B

Two conducting circular loops of radii $R _{1}$ and $R _{2}$ are placed in the same plane with their centres coinciding. If $R _{1} \gg R _{2}$, the mutual inductance $M$ between them will be directly proportional to

1. $R _{1}/R _{2}$

2. $R _{2}/R _{1}$

3. $R _{1}^{2}/R _{2}$

4. $R _{2}^{2}/R _{1}$

Correct Option: D

The mutual inductance $M _{12}$ of coil 1 with respect to coil 2

1. increases when they are bought nearer.

2. depends on the current passing through the coils.

3. increases when one of them is rotated about an axis.

4. is not same as $M _{21}$ of coil 2 with respect to coil 1.

Correct Option: A

A long solenoid  of diameter $0.1\ m$ has $2 \times {10^4}$ turns per metre.At the centre of the solenoid, a coil of $100$ turns and radius $0.01\ m$ is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to $0\ A$ from $4\ A$ in $0.05\ s$. If the resistance of the coil is $10 \ {\pi ^2}\Omega ,$ the total charge flowing through the coil during this time is.

1. $32\ \pi \mu C$

2. $16\ \mu C$

3. $32\ \mu C$

4. $16\ \pi \mu C$

Correct Option: C
Explanation:
Given,

Number of turns, $n=100$

Radius, $r=0.01\,m$

Resistance, $R=10\pi^2 \Omega$

As we know,

$\epsilon=-N\dfrac{d\phi}{dt}$

$=\dfrac{\epsilon}{R}=-\dfrac NR\dfrac{d\phi}{dt}$,   $\Delta I=-\dfrac NR\dfrac{d\phi}{dt}$

$\dfrac{\Delta}{\Delta t}=-\dfrac NR\dfrac{\Delta\phi}{\Delta t}\implies \Delta q=-[\dfrac NR(\dfrac{\Delta \phi}{\Delta t})]\Delta t$

$-$ve sign shoes that induced emf opposes the change in flux.

$\Delta q=\dfrac{\mu _0 ni\pi r^2}{R}$

$\Delta q=\dfrac{4\pi\times 10^{-7}\times 100\times 4\times \pi\times (0.01)^2}{10\pi^2}=32\mu C$

Two coils, a primary of $400$ turns and a secondary of $20$ turns are wound over an iron core of length $20\pi\ cm$ and cross-section of $2\ cm$ radius. If $\mu _{r}=800$, then the coefficient of mutual induction is approximately

1. $1.6\times 10^{7}H$

2. $1.6\times 10^{-2}H$

3. $1.6\times 10^{3}H$

4. $1.6\ H$

Correct Option: A

A charge of ${10^{ - 6}}C$ is describing a circular path of radius $1$ cm making $5$ revolution per second . The magnetic induction field at the centre of the circle is

1. $\pi \times {10^{ - 10}}T$

2. $\pi \times {10^{ - 9}}T$

3. $\frac{\pi }{2} \times {10^{ - 10}}T$

4. $\frac{\pi }{2} \times {10^{ - 9}}T$

Correct Option: D

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5\sin { \left( 10\pi t \right) }$ then the maximum value of e.m.f.induced in coil B is

1. $\pi \quad volt$

2. $\pi /2volt$

3. $\pi /3volt$

4. $\pi /4volt$

Correct Option: A

When the primary current in the spark-coil of a car changes from $4A$ to zero in $10\mu s$, an emf of $40000$V is induced in the secondary. The mutual inductance between the primary and the secondary winding of the spark-coil will be-

1. 1 H

2. 0.1 H

3. 10 H

4. zero

Correct Option: A

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5\sin { \left( 10\pi t \right) }$ then the maximum value of e.m.f. induced in coil B is

1. $\pi \quad volt$

2. $\pi /2volt$

3. $\pi /3volt$

4. $\pi /4volt$

Correct Option: A

The electric field of an electromagnetic wave is given by, $E=(50N^{-1})\, \sin { \omega } (t-x/c)$. Find the energy contained in a cylinder of cross section $10cm^2$ and length $50 cm$ along the x-axis.

1. $5.5\times 10^{-12}J$

2. $4.5\times 10^{-12}J$

3. $5\times 10^{-13}J$

4. $3.5\times 10^{-10}J$

Correct Option: A

An electron having kinetic energy T is moving in a circular orbit of radius R perpendicular to a uniform magnetic induction $\vec { \mathrm { B } }$  If kinetic energy is doubled and magnetic induction tripled, the radius will

1. $\frac { 3 R } { 2 }$

2. $\frac{{\sqrt 2 }}{3}R$

3. $\sqrt { \frac { 2 } { 9 } } R$

4. $\sqrt { \frac { 4 } { 3 } } R$

Correct Option: B
Explanation:

We know$,$

$R = \frac{{\sqrt {2mk} }}{{qB}}$
$R' = \frac{{\sqrt {2m2k} }}{{q\left( {3B} \right)}}$
$= \frac{{\sqrt 2 }}{3}\frac{{\sqrt {2mk} }}{{qB}}$
$= \frac{{\sqrt 2 }}{3}R$
Hence,
option $(B)$ is correct answer.

An average induced emf of 0.4 V appears in a coil when the current in it is changed from 10 A in one direction to 10 A in opposite direction in 0.5 sec. self-inductance of the coil is.

1. 0.008H

2. 50 H

3. 75 H

4. 100 H

Correct Option: A

At any instant t currenetI thorugh a coil of sself inductance 2mH is given.The induced e.m.f will be zero at time

1. 1 sec

2. 2 sec

3. 3 sec

4. 4 sec

Correct Option: B

When a current of 5 A flows in the primary coil then the flux linked with the secondary coil is 200 weber. The value of coefficient of mutual induction will be

1. 1000 H

2. 40 H

3. 195 H

4. 205 H

Correct Option: B
Explanation:

Coefficient of mutual induction will be the ratio of the flux linked with the secondary coil and the current primary coil.

$M = \dfrac{200}{5} = 40 H$

A cylindrical magnet is kept along the axis of a circular coil. On rotating the magnet about its axis. the coil will have induced in it

1. No current

2. A current

3. Only an e.m.f.

4. Both an e.m.f. and a current

Correct Option: A

A coil of insulating wire is connected to battery. If it is moved towards a galvanometer then its point gets deflected because

1. the coil behaves like a magnet

2. induced current is produced in the coil

3. the number of turns in the galvanometer coil remains constant

4. none of the above

Correct Option: B
Explanation:

Upon connecting to the battery , a current starts to flow in the insulated wire. This current induces a current in the galvanometer coil  due to mutual  inductance because of the flux linkages. This induced current causes a deflection in the galvanometer .

Two coils P and Q are lying parallels and very close to each other. Coil P is connected to an AC source whereas Q is connected to a sensitive galvanometer. On pressing key K

1. small variations are observed in the galvanometer for applied 50 Hz voltage

2. deflections in the galvanometer can be observed for applied voltage of 1 Hz to 2 Hz.

3. no deflection in the galvanometer will be observed

4. constant deflection will be observed in the galvanometer for 50 Hz supply voltage

Correct Option: B
Explanation:

Upon connecting to the AC source , a current starts to flow in the coil P. This current induces a current in the galvanometer coil Q due to mutual  inductance because of the flux linkages. This induced current causes a deflection in the galvanometer.

The value of mutual inductance can be increased by

1. decreasing N

2. increasing N

3. winding the coil on wooden frame

4. winding the coil on china clay

Correct Option: B
Explanation:

$M = \mu N _1 N _2 \dfrac{ Area _{12}}{length _{12}}$

$Area _{12}$ - area in common to both the coils where the flux links both of them together.
$length _{12}$ - length in common to both the coils where the flux links both of them together.
Therefore to increase the mutual inductance, the number of turns can be increased.

A long straight wire is placed along the axis of a circular ring of radius R. Then mutual inductance of this system is

1. $\frac{\mu _{0}R}{2}$

2. $\frac{\mu _{0}\pi R}{2}$

3. $\frac{\mu _{0}}{2}$

4. 0

Correct Option: D

A coil of wire has 0.2 m radius and 500 turns. It carries a current of 1 A.The magnetic induction at the centre of the coil is

1. $1.5\times { 10 }^{ -3 }T$

2. $1.6\times { 10 }^{ -3 }T$

3. $1.7\times { 10 }^{ -3 }T$

4. $1.8\times { 10 }^{ -3 }T$

Correct Option: B

The current in a coil is changed from 5 A to 10 A in $10^{-2}s$. Then, an emf of 50 m V is induced in a coil near by it. Calculate mutual inductance of two coils.

1. $100 \mu H$

2. $50 \mu H$

3. $20 \mu H$

4. $60 \mu H$

Correct Option: A

A short solenoid of radius  $a,$  number of turns per unit length  $n _ { 1 } ,$  and length  $L$  is kept coaxially inside a very long solenoid of radius  $b ,$  number of turns per unit length  $n _ { 2 } .$  What is the mutual inductance of the system?

1. $\mu _ { 0 } \pi b ^ { 2 } n _ { 1 } n _ { 2 } L$

2. $\mu _ { 0 } \pi a ^ { 2 } n _ { 1 } n _ { 2 } L ^ { 2 }$

3. $\mu _ { 0 } \pi a ^ { 2 } n _ { 1 } n _ { 2 } L$

4. $\mu _ { 0 } \pi b ^ { 2 } n _ { 1 } n _ { 2 } L ^ { 2 }$

Correct Option: C

A long straight wire is placed along the axis of a circular ring of radius $R$. The mutual inductance of this system is

1. $\cfrac{{\mu} _{0}R}{2}$

2. $\cfrac{{\mu} _{0}\pi R}{2}$

3. $\cfrac{{\mu} _{0}}{2}$

4. $0$

Correct Option: D

Two coils  $A$  and  $B$  having turns  $300$  and  $600$  respectively are placed near each other, on passing a current of  $3.0$  ampere in  $A$  the flux linked with  $A$  is  $1.2 \times 10 ^ { - 4 } weber$  and with  $B$  it is  $9.0 \times 10 ^ { - 5 } weber.$  The mutual  inductance of the system is

1. $2 \times 10 ^ { - 5 }$ henry

2. $3 \times 10 ^ { - 5 }$ henry

3. $4 \times 10 ^ { - 5 }$ henry

4. $6 \times 10 ^ { - 5 }$ henry

Correct Option: B

The inductance of a solenoid $0.5\ m$ long of cross-sectional area $20\ cm^2$ and with $500$ turns is

1. $12.5\ mH$

2. $1.25\ mH$

3. $15.0\ mH$

4. $0.12\ mH$

Correct Option: A

Two coils are placed close to each other. The mutual inductance of the pair of coils depend upon :

1. the currents in the two coils

2. the rates at which currents are changing in the two coils

3. relative position and orientation of the two coils

4. the materials of the wires of the coil

Correct Option: C
Explanation:

Mutual inductance between two coils is defined as the property of the coil due to which it opposes the change of current in the other coil. When the current in the neighboring coil changes, the flux sets up in the coil and because of this, changing flux emf is induced in the coil called Mutually Induced emf.

So, it depends on the relative position and orientation of two coils.

Which of the following units denotes the dimension $\dfrac {ML^2}{Q^2}$ where Q denotes the electric charge?

1. $Wb/m^2$

2. $henry (H)$

3. $H/m^2$

4. $weber (Wb)$

Correct Option: B
Explanation:
Mutual inductance $=\dfrac {\phi}{I}=\dfrac {BA}{I}$

$[Henry]=\dfrac {[MT^{-1}Q^{-1}L^2]}{[QT^{-1}]}=ML^2Q^{-2}$

The mutual inductance between two coils when a current of $5$ A changes to $10$ A in $1$ s and induces an emf of $100$ m V in the secondary is ______

1. $20$ m H

2. $10$ mH

3. $30$ mH

4. $15$ mH

Correct Option: A
Explanation:

Mutual inductance$=M=\dfrac{emf}{\Delta I/t}$

$=\dfrac{100mV}{10-5}$

$=20mH$

Identify which of the following best describe the Mutual inductance?

1. the ability of a current carrying conductor to induce a voltage in another conductor through a mutual magnetic field.

2. the ability of current carrying conductor to produce a changing magnetic field.

3. the ability of a conductor to induce a magnetic field in another current carrying conductor.

4. the ability of a current carrying conductor to induce a current in another conductor through a mutual magnetic field.

5. the ability of a magnetic field to induce a voltage in a current carrying conductor.

Correct Option: A
Explanation:

As per the Faraday's experiments, a current in a conductor produces a magnetic field. This magnetic field gets linked with another conductor and result in an induced emf.

For a current carrying inductor, emf associated in $20mV$. Now, current through it changes from $6A$ to $2A$ in $2s$. The coefficient of mutual inductance is

1. $20mH$

2. $10mH$

3. $1mH$

4. $2mH$

Correct Option: B
Explanation:

$\displaystyle \left | e \right |=L\frac{dI}{dt}$
Here, $\displaystyle e=20mV=20\times 10^{-3}V$
Coefficient of mutual inductance,
$20\times 10^{-3}=L\times 2$
$\displaystyle \therefore L=10\times 10^{-3}=10mH$

Two coils have a mutual inductance of $0.005\ H$. The current changes in the first coil according to equation $I=I _0sin\omega t$, where $I _0=10A$ and $\omega=100\pi rad/s$. The maximum value of emf (in volt) in the second coil is.

1. $2\pi$

2. $5\pi$

3. $\pi$

4. $4\pi$

Correct Option: B
Explanation:

$EMF=\frac {MdI}{dt}$
$=(0\cdot 005)I _0 w cos wt$
Maximum EMF$=(0\cdot 005)\times 10\times 100\pi$
$=5\pi$

The mutual inductance of the system of two coils is $5mH$. The current in the first coil varies according to the equation $I={ I } { o }\sin { wt }$ where ${ I } _{ o }=10A$ and $W=100\pi \, rad/s$. The value of maximum induced emf in the second coil is ______

1. $2\pi V$

2. $\pi V$

3. $5\pi V$

4. $4\pi V$

Correct Option: C
Explanation:

$Emf=M\cdot \cfrac { di }{ dt } =5\times { 10 }^{ -3 }\times { I } _{ o }\omega \cos { \omega t } \ { \left( Emf \right) } _{ max }=5\times { 10 }^{ -3 }\times { I } _{ o }\omega =5\times { 10 }^{ -3 }\times 10\times 100\pi =5\pi V$

A short solenoid of length $4cm$, radius $2cm$ and $100$ turns is placed inside and on the axis of a long solenoid of length $80cm$ and $1500$ turns. A current of $3A$ flows through the short solenoid. The mutual inductance of two solenoids is

1. $0.012H$

2. $5.3\times {10}^{-5}H$

3. $5.91\times {10}^{-3}H$

4. $8.3\times {10}^{-5}H$

Correct Option: C
Explanation:

As $M = \cfrac{\mu _0N _1N _2A}{l}$

where,
$A =$ common cross-sectional area
$l =$ length of small coil
$N _1 =$ No. of turns of small coil
$N _1 =$ No. of turns of long coil

$M = \cfrac{4\pi \times 10^{-7} \times 100 \times 1500 \times \pi \times (\cfrac{2}{100})^2}{(\cfrac{4}{100})} = 59157.6 \times 10^{-7} = 5.91 \times 10^{-3} H$

When the current in a coil changes from 8 ampere to 2 ampere in $3 \times 10^{-2}$ second, the e.m.f. induced in the coil is 2 volt. The self inductance of the coil (in millinery) is

1. 1

2. 5

3. 20

4. 10

Correct Option: A
Explanation:

$E.M.F. = L \dfrac{di}{dt}$

$2 = L \times \dfrac{8-2}{3 \times 10^{-2}}$

L = 1 millinery

Two coils have mutual inductance $0.005 H$. The current changes in the form coil according to equation, $I = I _0 \sin \omega t .$ Where $I _0 = 10 A.$ and $\omega = 100 \pi$ rads/s. The maximum value of emf in the second coil is :

1. $12 \pi$

2. $8 \pi$

3. $5 \pi$

4. $2 \pi$

Correct Option: C
Explanation:
Mutual inductance between two coils
M = 0.005 H
Peak current $l _0 = 10 A$
Angular frequency $\omega = 100 \pi$ rad/s
Current $l = l _0 \sin \omega t$
$\dfrac {d}{dt} = \dfrac {d}{dt} ( l \sin \omega t )$
$= l _0 \cos \omega t . \omega$
$= 10 \times 1 \times 100 \pi$
$= 1000 \pi$
Hence, induced emf is given by
$E = M \times \dfrac {dl}{dt}$
$= 0.005 \times 1000 \times \pi = 5 \pi V$

A coil of area 500 $cm^2$ having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude $4 \times 10^{-5}$ $weber/m^2$. If it is rotated by 180 about an axis passing through one of its diameter in 0.1 sec, find the average induced emf.

1. zero.

2. 30 mV

3. 40 mV

4. 50 mV

Correct Option: C
Explanation:
0iven that :-  $N=1000, B=4\times 10^{-5}weber/m^2, A=500cm^2=0.05m^2$

Initial flux linked with the coil, $\phi _1=1000\times 4\times 10^{-5}\times 0.05$

$\implies \phi _1=2\times 10^{-3}weber$

After rotation of $180^{o}$, B remains same but normal vector gets reversed, hence $\phi _2=-\phi _1$

Average EMF=$E=\dfrac{-\Delta \phi}{t}$

$\implies E=-\dfrac{\phi _2-\phi _1}{t}$

$\implies E=\dfrac{\phi _1-\phi _2}{t}$

$\implies E=\dfrac{2\phi _1}{t}$

$\implies E=\dfrac{4\times 10^{-3}}{0.1}V$

$\implies E=40mV$

A long straight wire is placed along the axis of a circular ring of radius $R$. The mutual inductance of this system is

1. $\dfrac{\mu _{0}R}{2}$

2. $\dfrac{\mu _{0}\pi R}{2}$

3. $\dfrac{\mu _{0}}{2}$

4. $0$

Correct Option: D

A coil of $Cu$ wire (radius $-r$, self-inductance-$L$) is bent in two concentric turns each having radius $\dfrac{r}{2}$. The self-inductance is now

1. $2L$

2. $L$

3. $4L$

4. $\dfrac{L}{2}$

Correct Option: C

Mutual inductance of a system of two thin coaxial conducting loops of radius 0.1 m, and then center separated by distance 10 m is (Take $\mu^{2} = 10$)

1. $2 \times 10^{-20}$ H

2. $2 \times 10^{-13}$ H

3. $2 \times 10^{-18}$ H

4. $2 \times 10^{-15}$ H

Correct Option: A

A ring of radius $r$ is uniformly charged with charge $q.$ If the ring is rotated about it's axis with angular frequency $\omega$, then the magnetic induction at its centre will be-

1. $10 ^ { - 7 } \times \frac { \omega } { q r }$

2. $10 ^ { - 7 } \times \frac { 9 } { \omega r }$

3. $10 ^ { - 7 } \times \frac { r } { q \omega }$

4. $10 ^ { - 7 } \times \frac { q \omega } { r }$

Correct Option: D
Explanation:

$\begin{array}{l} T=\dfrac { { 2\pi } }{ w } \ i=\dfrac { { qw } }{ { 2\pi } } \ B=\dfrac { { { \mu _{ 0 } }i } }{ { 2r } } =\dfrac { { { \mu _{ 0 } } } }{ { 2r } } \times \dfrac { { qw } }{ { 2\pi } } \ ={ 10^{ -7 } }\times \dfrac { { qw } }{ r } \ Hence, \ option\, \, D\, \, is\, correct\, \, answer. \end{array}$

Give the MKS units for the following quantities.
Magnetic Induction.

1. Weber$/m^3$.

2. Weber$/m^2$.

3. Weber$/m^4$.

4. Weber$/m^5$.

Correct Option: B
Explanation:

B. $Weber/m^2$

The term magnetic flux density refers to the fact that B is magnetic flux per unit surface. This relationship is based on Faraday's law of magnetic induction.
The SI unit measuring the strength of B is T(Tesla= $Weber/m^2$)

Two coaxial coils are very close to each other and their mutual inductance is $5mH$. If a current $50\sin{500t}$ is passed on one of the coils then the peak value of induced emf in the secondary coil will be:

1. $5000V$

2. $500V$

3. $150V$

4. $125V$

Correct Option: D
Explanation:

By the principle of mutual inductance we know that flux induced in coil2 will depend on the current flowing in coil1

$\phi _1=Mi _1$
Here
M=Mutual inductance
$i _1$=current in 1st coil
Now by Faraday's law induced emf in 2nd coil will be
$\frac { d\Phi }{ dt }=emf$
$emf=M\frac{di _1}{dt}$
Now put value of current and mutual inductance
$emf=(5\times 10^{-3}) \frac{d (50\sin500t)}{dt}$
$emf=(5\times 10^{-3})\times 50\times 500\times \cos500t$
$emf=125\cos500t$
So above is the equation of induced emf
So maximum value of induced emf will be
$emf _{max}=125 volts$

A solenoid would over a rectangular frame. If all the linear dimensions of the frame are increased by a factor $3$ and the number of turns per unit length remains the same, the inductance increased by a factor of :-

1. $3$

2. $9$

3. $27$

4. $63$

Correct Option: B
Explanation:

Firstly, the relation between inductance and length is calculated

$emf=\dfrac{-Nd\phi }{dt}=-NA\dfrac{dB}{dt}$

$B=\mu \dfrac{N}{l}I$

$emf=-\dfrac{\mu {{N}^{2}}A}{l}\dfrac{dI}{dt}$

$emf=-L\dfrac{dI}{dt}$

$L=\dfrac{\mu {{N}^{2}}A}{l}$

So, L is inversely proportional to length of solenoid.

If all the linear dimensions of the frame are increased by a factor 3 and the number of turns per unit length remains the same

Then area \begin{align} A=3l\times 3b  =9lb  {{L}^{'}}=\dfrac{9{{\mu }_{0}}{{N}^{2}}A}{l}  L'=9\,L $So new inductance becomes 9 times. Two different coils have self inductance$L _{1}=8\ mH, L _{2}=2\ mH$. The current in the second coil is also increased at the same constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are$i _{1}, V _{1}$and$W _{1}$respectively. Corresponding values for the second coil at the same instant are$i _{2}, V _{2}$and$W _{2}$respectively. Then 1.$\dfrac{i _{1}}{i _{2}}=\dfrac{1}{4}$2.$\dfrac{i _{1}}{i _{2}}=4$3.$\dfrac{W _{2}}{W _{1}}=4$4.$\dfrac{V _{2}}{V _{1}}=\dfrac{1}{4}$Correct Option: A,C,D Explanation: We know$e=L\dfrac{di}{dt}e\propto L$So,$\dfrac{e _1}{e _2}=\dfrac{L _1}{L _2}=\dfrac{8}{2}=\dfrac{4}{1}$Since$P=el=Constant$Therefore,$\dfrac{di _1}{dt}=\dfrac{di _2}{dt}P _1=P _2=Pe _1i _1=e _2i _2\therefore \dfrac{i _1}{i _2}=\dfrac{e _2}{e _1}=\dfrac{1}{4}$Thus, ratio of current is 1:4 Two coils A and B having turns 300 and 600 respectively are placed near each other, on passing a current of 3.0 ampere in A, the flux linked with A is 1.2 x$10^{-4}$weber and with B it is 9.0 x$10^{-5}$weber. The mutual induction of the system is: 1. 8 x$10^{-5}$H 2. 3 x$10^{-5}$H 3. 4 x$10^{-5}$H 4. 6 x$10^{-5}$H Correct Option: A Explanation: In passing current$(I=3A)$through coil$A$, the flux linked$(\phi _1)$is given by:$\phi _1=\mu _0 N _1^2I \left (\dfrac {A}{L}\right)\quad [N _1=300,\ given\\ A=Area\ of\ coil.\ L=Length]$and material inductance$(M)$between$A$&$B$given by:$M=\mu _0 N _1 N _2\left (\dfrac {A}{L}\right)$(Assuming both coils have same$A$&$L$) so from above,$M=\dfrac {N _2}{N _1}\dfrac {\phi _1}{I}=\dfrac {600}{300}\times \dfrac {1.2\times 10^{-4}}{3}\Rightarrow \ M=8\times 10^{-5}H$(Ans) Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area$A=10cm^2$and length$=20cm$. If one of the solenoids has$300$turns and the other$400$turns, their mutual inductance is$\left(\mu _o=4\pi\times 10^{-7} TmA^{-1}\right)$1.$4.8\pi \times 10^{-4}H$2.$4.8\pi \times 10^{-5}H$3.$2.4\pi \times 10^{-4}H$4.$2.4\pi \times 10^{-5}H$Correct Option: D Explanation:$\mu _{12}=\mu _on _1n _2lA4\pi \times 10^{-7}\times 300\times 400 \times 10\times 10^{-4}\times 20\times 10^{-2}=2.4\times 10^{-5}$A$50\ Hz$ac current of peak value$2$A flows through one of the pair of coils. If the mutual inductance between the pair of coils is$150\ mH$. then the peak value of voltage induced in the second coil is 1.$30\pi\ V$2.$60\pi\ V$3.$15\pi\ V$4.$300\pi\ V$5.$3\pi\ V$Correct Option: A Explanation: Let the current flows through coil 1 is,${{I} _{1}}={{I} _{0}}\sin \omega t$where,${{I} _{0}}$is peak value of current. Magnetic flux linked with coil 2 is${{\phi } _{2}}=M   {{I} _{1}}=M{{I} _{0}}\sin \omega t$Emf in coil 2 is${{\varepsilon } _{2}}=\dfrac{d{{\phi } _{2}}}{dt}=\dfrac{d(M{{I} _{0}}\sin \omega t)}{dt}=M \omega {{I} _{0}}\cos \omega t$So, peak value of voltage induced in coil 2 is$=M{{I} _{o}}\omega ....(1)$Given that,$ \nu =50\,Hz  {{I} _{0}}=2\,A  L=150\,mH  \omega =2\pi \nu =2\pi \times 50 $Put all values in equation (1)$ current=150\times {{10}^{-3}}\times 2\times 100\pi   I=30\pi \,V $The self-inductances of two identical coils are$0.1\ H$. They are wound over each other. Mutual inductance will be- 1.$0.1\ H$2.$0.2\ H$3.$0.01\ H$4.$0.05\ H$Correct Option: A The length of solenoid is 0.3 m and the number of turns is 2000. The area of cross-section of the solenoid is$1.2\times 10^{-3} m^2$. Another coil of turns 200 is wrapped over the solenoid. a current of 2 A is passed through the solenoid and its direction is changed in 0.25 sec. Then the induced emf in the coil: 1.$4.8\times10^{-2} V$2.$4.8\times10^{-3} V$3.$3.2\times10^{-3} V$4.$3.2\times10^{-2} V$Correct Option: A Explanation:$N _1=2000N _2=300A=1.2\times 10^{-3}m^2l=0.3\ mt=0.25\ sI=0.2\ A$we know, Mutual inductance$(M)=\dfrac{\mu _0N _1N _2A}{l}\Rightarrow M=\dfrac{\mu _0N _1N _2A}{l}$also, induced emf$(\varepsilon)=M.\dfrac{dI}{dt}$and$\dfrac{dI}{dt}=\dfrac{2-(-2)}{0.25}=\dfrac{4}{0.25}=16$so$\varepsilon =M\dfrac{dI}{dt}=\dfrac{\mu _0N _1N _2A}{l}.\dfrac{dI}{dt}\varepsilon=\dfrac{4\pi \times 10^{-7}\times 2000\times 300\times 1.2\times 10^{-3}\times 16}{0.3}\boxed{\varepsilon=0.048\ V=4.8\times 10^{-2}\ V}$hence$(A)$option is correct A coil of radius$1\ cm$and of turns$100$is placed in the middle of a long solenoid of radius$5\ cm$and having$5\ turns/ cm$. The mutual induction in millihenry is 1.$0.0316$2.$0.063$3.$0.105$4. Zero Correct Option: A When 100 volts d.c. is applied across solenoid a current of 1.0 amp flows in it. When 100 volts a.c. is applied across the same coil, the current drops to 0.5 amp. If the frequency of the a.c. source is 50 Hz the impedance and inductance of the solenoid are 1. 200 ohm and 0.55 Henry 2. 100 ohm and 0.86 Henry 3. 200 ohm and 1.0 Henry 4. 100 ohm and 0.93 Henry Correct Option: A The inductance of a solenoid 0.5 m long of cross-sectional area$420 cm^{2}$and with$500$turns is 1.$12.5\ mH$2.$1.25\ mH$3.$15.0\ mH$4.$0.12\ mH$Correct Option: B A current I flows in an infinity long wire with cross section in the from of a semicircular ring of radius R the magnitude of the magnetic induction along its axis is :- 1.$\dfrac { \mu _ { 0 } I } { 2 \pi R }$2.$\dfrac { \mu _ { 0 } \mathbf { I } } { 4 \pi R }$3.$\dfrac { \mu _ { 0 } I } { \pi ^ { 2 } R }$4.$\dfrac { \mu _ { 0 } I } { 2 \pi ^ { 2 } R }$Correct Option: C Explanation: Let semicircular ring of radius$R$as shown in figure An elementary length$dl$cut for finding magnetic field, So,$dl = Rd\theta $So, elementary current along$dl$is$di = \frac{i}{{\pi R}} \times Rd\theta  = \frac{{id\theta }}{\pi }$. Now, you can see that elementary part of length is perpendicular upon$dB$. so,$dB = \frac{{{\mu _0}di}}{{2\pi R}} = \frac{{{\mu _0}di}}{{2{\pi ^2}R}}$Now, magnetic filed along axis$B = \int\limits _0^\pi  {db.\sin \theta d\theta }  = 2dB$Now, put$dB$in here, So,$B = \frac{{{\mu _o}i}}{{{\pi ^2}R}}$Hence, Option$C$is the correct answer. What is inductance of a 25 cm long solenoid if it has 1000 turns an radius of its circular cross-section is 5 cm ? 1. 0.04 H 2. 0.02 H 3. 0.8 H 4. 0.1 H Correct Option: B A coil of mean area 500$cm^2$and having 1000 turns is held perpendicular to a uniform field of 0.4 gauss. The coil is turned through$180^o$in$\frac{1}{10}$second. The average induced e.m.f. :- 1.$0.04 V$2.$0.4 V$3.$4 V$4.$0.004 V$Correct Option: A The M.I. of a disc about its diameter is$2$units. Its M.I. about axis through a point on its rim in the plane of the disc is 1.$4$units 2.$6$units 3.$8$units 4.$10$units Correct Option: A A ring of radius r is uniformly charged with charge$q$. If the ring is rotated about it's axis with angular frequency$\omega$, then the magnetic induction at its centre will be- 1.$10 ^ { - 1 } \times \frac { \omega } { q r }$2.$10 ^ { - 7 } \times \frac { 9 } { \omega r }$3.$10 ^ { - 7 } \times \frac { r } { q \omega }$4.$10 ^ { - 7 } \times \frac { q \omega } { r }$Correct Option: D Explanation:$\begin{array}{l} T=\dfrac { { 2\pi  } }{ w }  \ i=\dfrac { { qw } }{ { 2\pi  } }  \ B=\dfrac { { { \mu _{ 0 } }i } }{ { 2r } } =\dfrac { { { \mu _{ 0 } } } }{ { 2r } } \times \dfrac { { qw } }{ { 2\pi  } }  \ ={ 10^{ -7 } }\times \dfrac { { qw } }{ r }  \ Hence, \ option\, \, D\, \, is\, correct\, \, answer. \end{array}$The self inductance of a coil having$500$turns is$50$mH. The magnetic flux through the cross-sectional area of the coil while current through it is$8$mA is found to be? 1.$4\times 10^{-4} Wb$2.$0.04$Wb 3.$4\mu$Wb 4.$40$m Wb Correct Option: A Explanation: Given that, Number of turns$N =500$Self inductance$L=50\times10^{-3}\ H$Current$I=8\times10^{-3}\ A$The magnetic flux through an inductor is the self inductance of coil times the current through it. The flux is$\phi=LI\phi=50\times10^{-3}\times8\times10^{-3}\phi=4\times10^{-4}\ Wb$So, the magnetic flux is$4\times10^{-4}\ Wb$Hence, A is correct option. What is the mutual inductance of coil and solenoid if a solenoid of length$0.50\ m$and with$5000$turns of wire has a radius$4\ cm$and a coil of$700$turns is wound on the middle part of the solenoid? 1.$44.17\ mH$2.$48.98\ mH$3.$34.34\ mH$4.$36.73\ mH$Correct Option: A The mutual inductance of an induction coil is 5 H. In the primary coil, the current reduces from 5 A to zero in$10^{-3} s$. What is the induced e.m.f. in the secondary coil? 1. 2500 V 2. 25000 V 3. 2510 V 4. zero Correct Option: B Explanation:$EMF=L\dfrac { di }{ dt } =5\times \dfrac { 5 }{ { 10 }^{ -3 } } =25000V$Two concentric rings are kept in the same plane. Number of turns in each rings is$25$. Their radii are$50 cm$and$200 cm$and they carry electric currents of$0.1 A$and$0.2 A$respectively, in mutually opposite directions. The magnitude of the magnetic field produced at their centre is ____________$T$. 1.$2{ \mu } _{ 0 }$2.$4{ \mu } _{ 0 }$3.$\dfrac { 10 }{ 4 } { \mu } _{ 0 }$4.$\dfrac { 5 }{ 4 } { \mu } _{ 0 }$Correct Option: D Explanation: Given,${ N } _{ 1 }={ N } _{ 2 }=25$turns${ R } _{ 1 }=50 cm=0.5 m{ R } _{ 2 }=200 cm=2 m{ i } _{ 1 }=0.1A, { i } _{ 2 }=0.2A$The magnitude of the magnetic field$\Delta B={ B } _{ 1 }-{ B } _{ 2 }=\dfrac { { \mu  } _{ 0 }{ N } _{ 1 }{ i } _{ 1 } }{ 2{ R } _{ 1 } } -\dfrac { { \mu  } _{ 0 }{ N } _{ 2 }{ i } _{ 2 } }{ 2{ R } _{ 2 } } =\dfrac { { \mu  } _{ 0 }\times 25 }{ 2 } \left( \dfrac { { i } _{ 1 } }{ { R } _{ 1 } } -\dfrac { { i } _{ 2 } }{ { R } _{ 2 } }  \right) =\dfrac { { \mu  } _{ 0 }\times 25 }{ 2 } \left( \dfrac { 0.1 }{ 0.5 } -\dfrac { 0.2 }{ 2 }  \right) =\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\left( \dfrac { 1 }{ 5 } -\dfrac { 1 }{ 10 }  \right) =\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\left( \dfrac { 2-1 }{ 10 }  \right)=\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\times \dfrac { 1 }{ 10 } =\dfrac { 25 }{ 20 } { \mu  } _{ 0 }=\dfrac { 5 }{ 4 } { \mu  } _{ 0 }$A solenoid of length 30 cm with 10 turns per centimetre and area of cross-Section 40$cm^2 $completely surrounds another co-axial solenoid of same length, area of Cross-section 20$cm^2$with 40 turns per centimetre. The mutual inductance of the 1. 10 H 2. 8 H 3. 3mH 4. 30 mH Correct Option: C Explanation: Given:$n _1 = 10cm^{-1} = 1000 m^{-1}n _2 = 40cm^{-1} = 4000 m^{-1}l= 30cm = 30 \, \times \, 10m^{-2}A _2 = 20cm^2 = 20 \, \times \, 10^{-4}m^2$Mutual inductance of the system,$M \, = \, \mu _0n _1n _2A _2l$(Where$A _2$is the area of inner solenoid.)$\therefore M = 4\pi  \times  10^{-7} \times  1000 \times 4000 \times 20 \times 10^{-4} \times 30 \times  10^{-2}M= 301.44 \, \times 10^{-5} H = 3mH$A 2 m long solenoid with diameter 2 cm an 2000 turns has a secondary coil of 1000 turns wound closely near its midpoint. The mutual inductance between the two coils is 1.$2.4\times 10^{-4}$H 2.$3.9\times 10^{-4}$H 3.$1.28\times 10^{-3}$H 4.$3.14\times 10^{-3}$H Correct Option: B Explanation: Here,$l = 2 m$, diameter = 2 cm$\therefore \, radius, \, r = \dfrac{2}{2} = 1cm = 1 \, \times \,10^{-2}mN _1$= 2000,$N _2$= 1000 Area =$\pi r^2$=$\pi \, \times \, \left ( 1\times 10^{-2} \right )^2 = 3.14 \times 10^{-4} m^2$Mutual inductance, M =$\dfrac{\mu _0N _1N _2A}{l}\,= \, \dfrac{4\pi \,\times \,10^{-7}\, \times \, 2000 \, \times \, 1000 \, \times \, 3.14 \,\times \,10^{-4}}{2}M=3.9 \, \times \, 10^{-4}H$A pair of adjacent coils has a mutual inductance of 2.5 H. If the current in one coil changes from 0 to 40 A in 0.8 s, then the change in flux linked with the other coil is then 1. 100 Wb 2. 120 Wb 3. 200 Wb 4. 250 Wb Correct Option: A Explanation: Mutual inductance of a pair of coils,,$M=2.5\ H$Initial current,$i _1=0\ A$Final current,$i _2=40\ A$Change in current,$di=i _2-i _1=40\ A$Time taken for the change,$t= 0.8\ sec$Induced e.m.f,$e= \dfrac{d\phi}{dt}= M\dfrac{di}{dt}$where$d\phi$is the change in the flux linked with the coil.$\implies d\phi = Mdi =2.5\times 40 =100\ Wb$Hence, the change in the flux linkage is$100\ Wb$. So, option$(A) is correct.

A short solenoid of radius a, number of turns per Unit length $n _1$. and length L is kept coaxially inside a very long solenoid of radius b, the number of turns per Unit length $n _2$. What is the mutual inductance of the system?

1. $\mu _0\pi b^2 n _1 n _2 L$

2. $\mu _0\pi a^2 n _1 n _2 L ^2$

3. $\mu _0\pi a^2 n _1 n _2 L$

4. $\mu _0\pi b^2 n _1 n _2 L ^2$

Correct Option: C
Explanation:

Let $L$ be the length of each solenoid $S _1$ and $S _2$ having radius a and b respectively.

$n _1$ and $n _2$ be the number of turns per unit length of $S _1$ and $S _2$.
And $I$ be the current through solenoid $S _2$
Magnetic field in $S _2= B _2= {\mu _0n _2I}$
Magnetic flux linked with each turn of $S _1 = B _2 \times$ area of each turn $= B _1\pi a^2$

Total magnetic flux linked with $S _1= B _2\pi a^2n _1L$

$\therefore \phi _1 = \left({\mu _0n _2I}\right)\pi a\ ^2n _2L = {\mu _0n _1n _2 \pi a^2 I}{L}$

But magnetic flux linked with $S _1$ is due to $I$
$\therefore \phi _1 \propto I$    or     $\phi _1 = M\ I$

Where $M$ is the mutual inductance of $S _2$ and $S _1$
$\therefore M \, I = {\mu _0n _1n _2\pi a^2I}{L}$

$\therefore M = \mu _0n _1n _2\pi a^2L$

Two short bar magnets of magnetic moment 'M' each are arranged at the opposite corners of a square of side 'o', such that their centres coincide with the square. If the like poles are in the same direction, the magnetic induction at any of the other of the square is

1. $\frac { { \mu } _{ 0 } }{ 4\pi } \frac { M }{ { d }^{ 3 } }$

2. $\frac { { \mu } _{ 0 } }{ 4\pi } \frac { 2M }{ { d }^{ 3 } }$

3. $\frac { { \mu } _{ 0 } }{ 2\pi } \frac { 3M }{ { d }^{ 3 } }$

4. $\frac { { \mu } _{ 0 } }{ 2\pi } \frac { 2M }{ { d }^{ 3 } }$

Correct Option: A

A conducting ring of radius r and resistance R rolls on a horizontal surface with constant velocity v.

1. The induced emf.between O and Q is 2 Bvr.

2. An induced current $I= \dfrac { 2Bvr }{ R }$ flows in the clockwise direction.

3. An induced current $I= \dfrac { 2Bvr }{ R }$ flows in the anticlockwise direction.

4. No current flows

Correct Option: D

Two coils of self-inductances $2\ mH$ and $8\ mH$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

1. $10\ mH$

2. $6\ mH$

3. $4\ mH$

4. $16\ mH$

Correct Option: C
Explanation:

Mutual inductance M=$\sqrt{ L _1\times L _2}=\sqrt{ 2\times 8}=4$mH

Mutual inductance of a system of two thin coaxial conducting loops of radius each, their centers separated by distance $d (d >>r)$ is

1. $\mu _0\pi r^4d^3$

2. $\dfrac{\mu _0\pi r^4}{2d^3}$

3. $\dfrac{\mu _0\pi r^4}{d^3}$

4. $\dfrac{\mu _0\pi r^4 d^3}{4}$

Correct Option: B

The coefficient of mutual inductance, when magnetic flux changes by $\displaystyle 2\times { 10 }^{ -2 }Wb$ and current changes by 0.01 A is :

1. 8 henry

2. 4 henry

3. 3 henry

4. 2 henry

Correct Option: D
Explanation:

We know that

$\displaystyle \phi =Mi$

$\displaystyle d\phi =Mdi$

$\displaystyle M=\frac { d\phi }{ di } =\frac { 2\times { 10 }^{ -2 } }{ 1\times { 10 }^{ -2 } } =2$ henry

- Hide questions


text:
# Multicomplex number

In mathematics, the multicomplex number systems Cn are defined inductively as follows: Let C0 be the real number system. For every n > 0 let in be a square root of −1, that is, an imaginary number. Then $\text{C}_{n+1} = \lbrace z = x + y i_{n+1} : x,y \in \text{C}_n \rbrace$. In the multicomplex number systems one also requires that $i_n i_m = i_m i_n$ (commutativity). Then C1 is the complex number system, C2 is the bicomplex number system, C3 is the tricomplex number system of Corrado Segre, and Cn is the multicomplex number system of order n.
The multicomplex number systems are not to be confused with Clifford numbers (elements of a Clifford algebra), since Clifford's square roots of −1 anti-commute ($i_n i_m + i_m i_n = 0$ when mn for Clifford).

format: plain


text:
Advertisement
11,354
pages

Tredecation is the 13th hyperoperation. It is equal to the binary function $$\uparrow^{11}$$ in Knuth's up-arrow notation[1]

The n tredecated to m is n doedecated to n doedecated to ... doedecated to n (m n's).

"a tredecated to b" can be written in array notation as $$\{a,b,11\}$$, in chained arrow notation as $$a \rightarrow b \rightarrow 11$$ and in Hyper-E notation as E[a]1#1#1#1#1#1#1#1#1#1#1#1#b.

Tredecational growth rate is approximately equivalent to $$f_{12}(n)$$ in the fast-growing hierarchy.

Advertisement


text:
(forgot?)
F-IF Influenza epidemic

### Alignments to Content Standards

• Functions
Domain
HSF-IF: Interpreting Functions
Cluster
Interpret functions that arise in applications in terms of the context.
Standard
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.

### Tags

This task is not yet tagged.

An epidemic of influenza spreads through a city. The figure below is the graph of $I=f(w)$, where $I$ is the number of individuals (in thousands) infected $w$ weeks after the epidemic begins.

1. Estimate $f(2)$ and explain its meaning in terms of the epidemic.
2. Approximately how many people were infected at the height of the epidemic? When did that occur? Write your answer in the form $f(a)=b$.
3. For approximately which $w$ is $f(w)=4.5$; explain what the estimates mean in terms of the epidemic.
4. An equation for the function used to plot the image above is $f(w)=6w(1.3)^{-w}$. Use the graph to estimate the solution of the inequality $6w(1.3)^{-w}\geq 6$. Explain what the solution means in terms of the epidemic.
5. (Task from Functions Modeling Change: A Preparation for Calculus, Connally et al., Wiley 2010.)

### Commentary

The principal purpose of this task is to probe students' ability to correlate symbolic statements about a function using function notation with a graph of the function, and to interpret their answers in terms of the quantities between which the function describes a relationship. It can be used in assessment, or in instruction to bring out some common frailties of student understanding, such as not really understanding what it means for a point to lie on the graph of a function, and, in part (d), not being comfortable with interchanging a function value expressed in function notation and an expression for the function. As it is impossible to read exact numerical data from the graph, students will have to approximate coordinates of data points on the graph, providing a good opportunity for an instructor to address MP6 (attend to precision).

#### Solutions

Solution: Influenza epidemic
1. To evaluate $f(2)$, we determine which value of $I$ corresponds to $w=2$. Looking at the graph, we see that $I\approx 7$ when $w=2$. This means that approximately $7000$ people were infected two weeks after the epidemic began.
2. The height of the epidemic occurred when the largest number of people were infected. To find this, we look on the graph to find the largest value of $I$, which seems to be approximately $8.5$, or $8500$ people. This seems to have occurred when $w=4$, or four weeks after the epidemic began. We can say that the height of the epidemic corresponds to the evaluation $f(4)=8.5$.
3. To find a solution to $f(w)=4.5$, we must find the value of $w$ for which $I=4.5$, or $4500$ people were infected. We see from the graph that there are actually two values of $w$ at which $I=4.5$, namely $w\approx 1$ and $w\approx 10$. This means that $4500$ people were infected after the first week when the epidemic was on the rise, and that after the tenth week, when the epidemic was slowing, $4500$ people remained infected.
4. We are looking for all the values of $w$ for which $f(w)\geq6$. Looking at the graph, this seems to happen for all values of $w\geq1.5$ and $w\leq8$. This means that more than $6000$ people were infected starting in the middle of the second week and lasting until the end of the eighth week, after which time the number of infected people fell below $6000$.

Comments entered here will be visible to the general public!

• BooDrury wrote this public comment 12 months ago

As stated above: "The purpose of this task is to probe students' ability to correlate symbolic statements about a function using function notation with a graph of the function, and to interpret their answers in terms of the quantities between which the function describes a relationship."

What exactly is the point of having even numbering scaling? The idea is to test function notation and not graphing interpretation, such as 1.5 weeks. Interpret above is used in context of function in-puts and out-puts not graphical scaling.

Am I missing some point?

• Cam McLeman wrote this public reply 11 months ago

Hi BooDrury,


text:
# Surface integral

1. Mar 5, 2014

### Jhenrique

The line integral can be expressed, at least, in this three different ways: $$\int \vec{f} \cdot \hat{t} ds = \int \vec{f} \cdot d\vec{s} = \int \vec{f} \cdot d\vec{r}$$ The surface integral too (except by least expression above): $$\iint \vec{f} \cdot \hat{n} d^2S = \iint \vec{f} \cdot d^2\vec{S}$$ My ask is: exist some equivalent/analogous expression to ∫ f·dr for surface integral? In other words, is possible to write the surface integral in terms of the position vector r? Maybe: $\iint \vec{f} \cdot d^2\vec{r}$ ?

2. Mar 9, 2014

### Einj

I don't really agree with the third formula you used for the line integral. The second one is definitely correct, the line integral takes the component of the vector field parallel to the line itself. The third one is not always valid. It is true in the case of radial motion, when you path is actually parallel to the position vector itself.

Last edited: Mar 9, 2014
3. Mar 9, 2014

### Jhenrique

Great observation! But, are you sure???? I already saw the 3rd definition a lot times.

https://en.wikipedia.org/wiki/Line_integral

4. Mar 10, 2014

### Einj

If you look at the Wikipedia page you linked it says that r(t) is a bijective parametrization of the curve, not the position vector. It's just a confusing notation, that why one usually uses s(t).

Inviato dal mio iPhone utilizzando Physics Forums

5. Mar 10, 2014

### Bill_K

r(t) doesn't have to be the position vector, the Wikipedia page is saying it can be more general, but it CAN be the position vector. For a line integral, you want the differential to be a vector parallel to the curve whose magnitude is the line element, and that's exactly what dr is. Equivalently, one can write dr = (dr/dt) dt.

For the surface integral, you want the differential to be a vector orthogonal to the surface whose magnitude is equal to the element of surface area. Suppose the surface is described by giving the position vector as a function of two variables, r(s, t). Then the differential is the vector d2S = (∂r/∂s) x (∂r/∂t) ds dt.


text:
# Modular arithmetic ; Can I please get help with this math problem [duplicate]

I'm very confused at how this math problem is being set up... I just came across a page on my study on congruences. I've studied the theorems on it but I seem to be lagging behind in this direction. Can I get help with these two questions and I would love to know more about the answer with steps to how the question was solved. Thanks in advance.

(1) Find $$2^{501} \pmod {17}$$

(2) Solve for $$41x \equiv 5\pmod {51}$$

• Question $(1)$ asks you to find the remainder when $2^{501}$ is divided by $17$. $(mod\ n)$ is an environment where you denote any integer $m$ by it's remainder when divided by $n$, so that if $m_1,m_2$ have same remainders on division by $n$, they are *equal*(or congruent, like geometric congruency) modulo n i.e. equal in the (mod n) environment. Where are you studying from? Do you know properties like $a\equiv b (mod \ n) \implies a^k\equiv b^k (mod\ n)$ for any positive integer $k$? – Fawkes4494d3 Aug 30 '20 at 1:01
• I know that... I'm actually studying from a document and the property you stated is in the document. – Dave Kent Aug 30 '20 at 1:15
• for (2), $x\equiv \frac{5}{-10} \equiv \frac{-1}{2} \equiv \frac{50}{2} \equiv 25 \pmod{51}$ – Evariste Aug 30 '20 at 1:37

For (1), note that $$2^4\equiv-1\bmod17$$, so $$(2^4)^{125}\equiv-1\bmod17$$.

For (2), use the extended Euclidean algorithm:

$$51=41+10$$

$$41=4\times10+1.$$

Therefore $$1=41-4\times10=41-4(51-41)=5\times41-4\times51$$,

so $$5=25\times41-20\times51$$.

Can you take it from here?


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

In my last blog post, I talked about Random Sampling for Plain Text in R. Now when you are performing some supervised learning on your data, if you follow the cross-validation principle which divides the data into train, holdout and test, you want to make sure each subset is different. The great caret package has some neat partitioning techniques however for my use, I wanted a minimal function without the need to load extra libraries.

Solution
createPartition <- function (filename, trainPortion=0.6, holdOutPortion=0.2) {
con <- file(filename)

sample <- floor(total * trainPortion)
sample_holdout <- floor(total * holdOutPortion)

corpus <- scan(con, what="character", skip= 0, nlines=total, sep="\n", fileEncoding = 'UTF-8')
close(con)
train <- corpus[1:sample]
holdout <- corpus[(sample + 1): (sample + sample_holdout)]
test <- corpus[(sample + sample_holdout + 1) : total]

result <- list(train=train, holdout=holdout, test=test)
return(result)
}

What you need to pass is only a path to filename. I set the partitions to be 60/20/20 rule which is suggested by the great Andrew Ng.

Here's how to use it:

data <- createPartition("path_to_file.format")
data$train data$holdout
data\$test

I hope this helped! ???


text:
See on GitHub

# Vyxal - Terse, Elegant and Readable

Vyxal is the latest addition to the plethora of stack-based languages used for code golfing. But unlike its competitors, Vyxal has a special focus upon readability and elegancy. Indeed, the codepage has been specially chosen to be as mnemonic as possible. Further, constructs from practical languages (such as functions, lambdas and easy list manipulation) are present.

(Vyxal is pronounced Vikesal)

## How to use the interpreter:

python3 Vyxal.py <file> <flags (single string of flags)> <input(s)>

For a list of command-line flags: python3 Vyxal.py h

## Data Types

There are 5 data types supported by Vyxal:

• Numbers (integers and reals/floats)
• Strings
• Lists
• Generators
• Functions

## Basic Operators

• +-*/% perform addition, subtraction, multiplication, division and modulo respectively.
• , prints the top of the stack
• :_ duplicates the top of the stack and pops the top of the stack respectively
• ! pushes the length of the stack

## Syntax Constructs

### If statements

[truthy_branch|falsey_branch]
[truthy_branch]

The if statement pops the top of the stack, and if it is truthy, executes the truthy branch. Otherwise, if a falsey branch is present, it will branch to execute that.

### For loop

(variable|body)
(body)

The for loop pops the top of the stack and iterates through each item. If the value popped is an integer, it loops through the range [0, n). If variable is present, the iteration value is stored in that. Otherwise, the iteration value is stored in the context variable n.

### While loop

{condition|body}
{body}

The while loop repeats body until condition evaluates as true. If there is no explicit condition, 1 is used as the condition, meaning that {...} is an infinite loop.

### Functions

@name|code;
@name:number_of_arguments|code;
@name:variable|code;
@name:argument_list|code;
@name;

If code isn’t present, the function with name name is called. Otherwise, the function is defined. The arguments can be a combination of variables and numbers. Numbers tell the function how many items to pop from the main stack as arguments, and variables store a single value in the variable. Numbered arguments are pushed to the function’s stack – functions operate on their own scoped stack with scoped variables (much like Python).

For example:

@triple:1|3*;

Takes 1 parameter and pushes it to the function’s stack

@triple:value|←value 3*;

Takes a single argument and stores it in variable value.

@add_and_halve:1:rhs|←rhs +2/;

Takes two arguments: pushes the first on to the stack and stores the second in variable rhs

@average:*|W:L$∑$/;

Takes however many arguments as defined by the first value popped from the stack. A function call of 2 3 3 3 @average; would take three arguments.

### Lambdas

λarity|code;
λcode;

Where the @...; function stores the definition for infinite re-use, the lambda pushes a reference to the code inside it. This is similar to python’s lambdas, which are temporary functions, or literal functions (for lack of better word).

These can be applied using †. For example: 3 λ3*; † will result in 9. Lambdas are also useful for mapping/filtering/reducing a vector according to the lambda’s code.

### Implicit input and output

• At the end of program execution (eof), if nothing has been printed (using , or other printing commands), the top value on the stack is automatically printed.
• If there isn’t enough values on the stack to perform an operation, implicit input is taken. If input is passed through command line arguments, then the input used is cycled.
• Input can be either through arguments or STDIN. STDIN is used if arguments aren’t avaliable. If no input is avaliable at all, 0 is returned.
• In functions (and lambdas), if implicit input is needed, the argument(s) passed are used as the input “list”.

### Commands

Vyxal has so many commands that it is impractical to list them all here. Here is the reference page (markdown).

## Examples

### Hello, World!

Hello, World!

Try it Online!

ƈṡ, ƛ€!

Try it Online!

The above program uses dictionary compression: words in a predefined list are indexed using a subjective base-162 literal.

kH

Try it Online!

### Fizzbuzz

₁ƛ₍₃₅kF½*∑⟇

### Prime Checking

æ

Try it Online!

KL2=
`

Try it Online!

## Very Special Contributors (alphabetically sorted)

• code-golf se user @2x-1 for helping me establish the fundamentals of Vyxal and being my first collaborator on this journey.
• @8dion8 for language suggestions and motivation in the MAWP discord group.
• @Amiller42 for pointing out bugs and fixing bugfixes
• @chunkybanana for also bug fixing and making PRs
• @detectivewasif for adding detail to the reference sheet.
• @Razetime for helping me with the online interpreter’s design
• @ysthakur for making the reference.md file and making an automated process to do so.


text:
# Cross Products isn't giving the right answer to angle.

Vectors in geometry

### Cross Products isn't giving the right answer to angle.

When I use cross products to find an angle of 2 vectors as compared to using dot products, I realised that it's wrong.

Using dot products to find angle
===================================
V = (-3, 1, 0)
W = (1,2,0)

Formula: cosΘ = V dot W /( |V||W| )

=> cosΘ = -1 / √50
=> cosΘ = -0.141421

Inverse of cos gives 98.13° which is the correct angle.

Using cross products to find angle
===================================
V = (-3, 1, 0)
W = (1,2,0)

Formula: |VxW| = |V||W|sinΘ

=> sinΘ = |VxW| / |V||W|
=> sinΘ = 7 / √50
=> sinΘ = 0.989949

Inverse of sin gives 81.869898° which is a wrong angle.

But if 180°-81.869898° gives the right answer.

Why is that so? Is it something wrong with the formula?

Posts: 1
Joined: Mon Jan 14, 2008 6:19 am
Reputation: 0

Or write them here.

Here is 2 good articles:
http://www.math10.com/en/geometry/vecto ... ctors.html
http://www.math10.com/en/geometry/vecto ... tions.html

Look at the second carefully, please!

Math Tutor

Posts: 418
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 32

### Re: Cross Products isn't giving the right answer to angle.

There is nothing wrong with the formula, gut you need to understand that the trig functions are not "one-to-one" so do not have true inverses. In particular, if x is a small number, sin(90+ x)= sin(90- x). Calculators typically give angles between -90 and 90 degrees ($$-\pi/2$$ to [/tex]\pi/2[/tex] radians) while it is clear that this answer is larger 90 degrees. It is not that the formula is wrong- you are using the calculator incorrectly.
Guest


text:
## FANDOM

162 Pages

Heisenberg's uncertainty principle or just the uncertainty principle is a principle in quantum physics which states that there is a limit to how precisely we can measure a particle's position and momentum at a given time. Mathematically, it is stated as

$\Delta x \Delta p \geq \frac{\hbar}{2} ~~$

where ħ is the reduced Planck constant (or the Planck constant over 2π).

Community content is available under CC-BY-SA unless otherwise noted.


text: So after struggling for a couple of hours, I figured it out. The main issue was that I did not have access to the source code of the demo, but noticed using rostopic that turtle_tf2_message_broadcaster.py was publishing the PointStamped message with "/world" frame. I had to remove the command that runs that script from the launch file of the demo, implemented my own turtle_tf2_message_broadcaster.py without the '/' in the "world" frame, and then everything worked beautifully.


text:
## Icon Index Symbol • 5

### Questions Concerning Certain Faculties Claimed For Signs

Re: Peirce List Discussion • Helmut Raulien

Given that sign relations are special cases of triadic relations, we can get significant insight into the structures of both cases by examining a few simple examples of triadic relations, without getting distracted by all the extra features that come into play with sign relations.

When I’m talking about a $k$-place relation $L$ I’ll always be thinking about a set of $k$-tuples.  Each $k$-tuple has the form:

$(x_1, x_2, \ldots, x_{k-1}, x_k),$

or, as Peirce often wrote them:

$x_1 : x_2 : \ldots : x_{k-1} : x_k.$

Of course, $L$ could be a set of one $k$-tuple but that would be counted a trivial case.

That sums up the extensional view of $k$-place relations, so far as we need it for now.

Using a single letter like $``L"$ to refer to a set of $k$-tuples is already the genesis of an intensional view, since we now think of the elements of $L$ as having some property in common, even if it’s only their membership in $L.$  When we turn to devising some sort of formalism for working with relations in general, whether it’s an algebra, logical calculus, or graph-theoretic notation, it’s in the nature of the task to “unify the manifold”, to represent a many as a one, to express a set of many tuples by means of a single sign.  That can be a great convenience, producing formalisms of significant power, but failing to discern the many in the one can lead to no end of confusion.

To be continued …

### 2 Responses to Icon Index Symbol • 5

This site uses Akismet to reduce spam. Learn how your comment data is processed.


text:
# Importance of Tracelessness of Tensor?

What makes the trace-free tensor (or part of it) so important? As in trace-free Ricci tensor or Weyl tensor.

Symmetry. A general rank-2 tensor $$T\in V\otimes V$$ can be decomposed as
$$T = \tau I + S + A,$$
where $$n\tau = \operatorname{Tr}(T)$$ ($$n$$ being the dimension of $$V$$), $$I$$ is the identity, $$S$$ is trace-less symmetric and $$A$$ is skew-symmetric. These "components" obey different transformation laws under a given symmetry. For example, the scalar part $$\tau I$$ is obviously invariant.