Homework Help: Magnetic fields and wire current

1. Apr 29, 2008

xeannart

Magnetic fields (urgent)

Hello, i'm in trouble, I have no idea what to do/look for the formula to solve these 2 questions.

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A straight wire has a current of 15 A vertically upwards, in a vacuum. An electron, presently 0.10 m from the wire moves at a speed of 5.0 x 106 m/s. Its instantaneous velocity is parallel to the wire but downward. Calculate the magnitude and direction of the force on the electron. Will this force remain constant?

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A singly ionized atom of mass 3.9 x 10-25 kg is accelerated through a potential difference of 1.0 x 105 V.

a) Calculate its maximum speed.
b) What is the radius of the path it would take if injected at this speed and at 90 degrees into a uniform magnetic field of magnitude 0.10 T?

2. Apr 29, 2008

rock.freak667

What you need here is to find the force exerted on the electron which is given by? (Hint: Force exerted on a charged particle(Q) moving with a velocity(v) in a magnetic field(B))
What is the formula for the magnetic field around a long straight current carrying conductor?
For this one...All the energy supplied by the p.d. goes into increasing the kinetic energy of the ions. (I am assuming that this is a mass spec. thing)

3. Apr 29, 2008

xeannart

For the first question it's these 2 equations right?
B = m(I/(2pi*r))
F = qVB*Sin(90)

I can plug the values in, but i'm missing mu (m). can anyone tell me it?

For the 2nd question, it's the centrepedial force equals the force of the electrons right?

qVB sin(90) = mv^2/r

4. Apr 29, 2008

Gear300

mu is 4*pi*10^-7.

For the second one, you seem to be on track with it.

5. Apr 29, 2008

Dozent100

All of which is in your text!
Edmund

6. Apr 30, 2008

xeannart

So for the first question the math is like this

B = m(I/(2pi*r))
B = 4*pi*e-7(15/(2pi*0.1)) = 3e-5
F = qVB*Sin(90)
F = (1.6e-19)(5e6)(3e-5)(1) = 2.4e-17 N

And so if upward is positive, this would point upward?

For the second question, the math would be like this?
a)
(1/2)mv^2 = Vq
v = 2.86e5

b)
qVB sin(90) = mv^2/r
qVB = mv^2/r
r = (mv^2)/(qVB)
r = (3.9e-25*(2.86e5^2))/((1.6e-19)(1e5)(0.1)) = 19.93 m

Oh i was wondering is mu a constant? where does this number (4*pi*e-7) come from?
V in qVB is speed right?

Last edited: Apr 30, 2008
7. Apr 30, 2008

Nick89

$\mu_0$ is called the permeability of free space, a defined constant.
It is not measured, it's defined as: $\mu_0 = 4 \pi \times 10^{-7}$

It is also related to the vacuum permittivity $\epsilon_0$ and the speed of light in a vacuum $c_0$:
$$c_0^2 \mu_0 \epsilon_0 = 1$$

8. Apr 30, 2008

xeannart

ahh i c. thx

for question 1, do we need to find the values for V using qVB=(1/2)mv^2/r before pluging it in to F = qVB??