# Homework Help: Force on electron as it travels through conductor

1. Jul 16, 2008

### NATSALANE

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

In a vacuum, a straight conductor has 18-A current that goes upward. An electron is traveling at 8.92 x 10^4 m/s. If the electron is 0.2 m from the conductor and its instantaneous velocity is parallel to the conductor (albeit heading downward), find the magnitude and direction of the force on the electron. Explain whether the force will be constant.

Here is my attempt:

Known:
I (current) = 18 A [upward]
v (speed) = 8.92 x 10^4 m/s
q (charge) = 1.602 x 10^-19 C

Unknown:
Fe

F magnetic = F centripetal

qvB = (m x v^2) / r

Therefore,
B = mv / qr
= (9.11 x 10^-31) x (8.92 x 10^4) / (1.602 x 10^-19) x (0.20)
= 0.000002536 T

F magnetic = qvB sin$$Theta$$
= (1.602 x 10^-19) x (8.92 x 10^4) x 0.000002536 x sin90
= 3.6 x 10^-20 N

I think it went horribly, horribly wrong. Was I supposed to use to F magnetic equation at all? It's difficult to know which F equation to use.

I'd appreciate any guidance with this problem. Thank you.

2. Jul 17, 2008

### Dick

It did go horribly wrong. The only field outside of the conductor is magnetic, but you can't assume that the electron is circling the conductor. It isn't. So your determination of B is wrong. You need to use Ampere's law to find the magnetic field around the conductor not 'centripetal force'. Once you understand what the magnetic field looks like, then yes, use q*(v x B) to find the force.

3. Jul 17, 2008

### NATSALANE

Thank you for your help. This is what I arrived at:

I (current) = 18 A [upward]
v (speed) = 8.92 x 10^4 m/s
q (charge) = 1.602 x 10^-19 C

B = μ0 [I/(2πr))
= 4π x 10^-7 [18 / (2 x π x 0.20)
= 1.8 x 10^-5

F nag. = qvB sinΘ
= (1.602x10^-19) x (8.92 x 10^4) x (1.8 x 10^-5) x sin90
= 2.57 x 10^-19 N

4. Jul 17, 2008

### Dick

That's the magnitude of the force alright. How about direction?

5. Jul 17, 2008

### NATSALANE

This is another part where I am stuck. Since the magnetic force is perpendicular to both the velocity and the magnetic field B, would the direction of the force be towards the observer?

6. Jul 17, 2008

### Dick

Toward the observer?? That can't be right. The observer could be anywhere. There are things called 'right hand rules' for remembering the relation between current flow and B field direction, and the directions of F and vxB. Look them up and try it out.

7. Jul 17, 2008

### NATSALANE

Using the right-hand rule, the thumb should be pointing up. Not sure which way to point the fingers, though... Would the direction of magnetic force be westward?

8. Jul 17, 2008

### Dick

Westward?? You need a coordinate system. Suppose j points in the direction of current flow (so -j is the direction of v). Suppose i is the direction of a vector pointing from the axis of the conductor towards the electron. Now complete the system by picking the direction k using the right hand rule. Now describe directions in terms of i,j and k.

9. Jul 18, 2008

### NATSALANE

The direction is horizontal to conductor.

If that's right, maybe you can help me with another question:

Two sources that are placed 2.0 m apart operate at a frequency of 1.0 Hz. If the waves are 0.60 m, at what angles (from the centre line of the interference pattern) are nodal lines produced?

Known:
λ = 0.6 m
d = 2.0m
f = 1.0 Hz

Unknown:
angle θ

sin θ = (n – ½) x (λ /d)
= (1-½) x (0.6/2)
= 8.63°

Therefore, the nodal lines are produced 8.6° from the centre line of the interference pattern.

10. Jul 18, 2008

### Dick

I don't know what direction "horizontal to conductor". The new question is entirely unrelated. I suggest you put it in a new thread.