# Homework Help: Magnetic force on a rigid semicircular wire

1. Jun 23, 2009

### Maxwellkid

1. The problem statement, all variables and given/known data
A rigid wire carrying a current I, consists of a semicircle of radius R and two straight portions of length L. The wire lies in in a plane perpendicular to a uniform magnetic field B. The straight portions are also within the magnetic field. Determine the net force due to B on the wire.

2. Relevant equations
F1 = IL1 x B
F2 = IL2 x B
F3 = dF = I dL3 x B

3. The attempt at a solution

can somebody help me start this problem off correctly by converting these vectors into a uniform coordinate system so that I cab take this simple integral?

B = <0, 0, -Zb>

Last edited: Jun 23, 2009
2. Jun 23, 2009

### LowlyPion

Welcome to PF.

Consider that the plane of the loop is perpendicular to the magnetic field.

Where will the forces from the B field necessarily be, that is in what direction?

3. Jun 23, 2009

### Maxwellkid

The forces for the two straight portions are easy.

F1 = <0, IL, 0> x <0, 0, -zB> = -IL(zB) i
F2 = <0, -IL, 0> x <0, 0, -zB> = IL(zB) i

F3 = dF3 = I dL x B
this is where I am stuck!!!

4. Jun 23, 2009

### Maxwellkid

Also, do magnetic fields of north or south pole attract or repel either a positive or negative charge??

5. Jun 23, 2009

### LowlyPion

Is the wire in a closed loop?

Or is it shaped like the outline of a derby hat?

I don't have the advantage of seeing any diagram.

6. Jun 23, 2009

### Maxwellkid

the rigid wire goes like this. straight up...then forms a semi circular curve...then straight down... no closure stated.

7. Jun 23, 2009

### LowlyPion

OK. So what direction are the forces on each of the 2 || segments?

What happens when you add them together?

8. Jun 23, 2009

### Maxwellkid

the two parallel section forces cancel each other out. I'm not sure about the upward force. I must take the integral, but I'm a bit stuck.

9. Jun 23, 2009

### LowlyPion

What upward force are you considering?

10. Jun 23, 2009

### Maxwellkid

the upward force due to the semi circle that connects the 2 parallel sections of the overall wire. F3

11. Jun 23, 2009

### LowlyPion

So long as you mean upward in the y direction and not out of the paper in z , as there is no force in the || direction to the B field. The cross product will only yield x,y forces. (The plane of your wire.)

Since the semi-circle is symmetrical, then you can ignore any x-directed forces, taking x to be perpendicular to the straight wires.

So maybe set up an integral of just the y-component recognizing that you can express the Force component in y as R*sinθ, where θ is the angle with the x axis. From 0 to π, since it is a semi-circle should do it don't you think?

12. Jun 23, 2009

### Maxwellkid

that's what i've been trying to do...

dF3 = IdL x B how does the integral of dF3 equal to the integral of Rsin@ d@?????

13. Jun 24, 2009

### LowlyPion

Note that your F3 can be written as

dF3(θ) = |dF3|cosθ i + |dF3|sinθ j

but by symmetry for each positive directed |dF3|cosθ i you have on the other side of the mid-line a negative |dF3|cosθ i component. This resolves itself into

dF3(θ) = |dF3|sinθ j

F3(θ) = ∫ |dF3|sinθ dθ , evaluated from 0 to π

This force is all y-directed, since all x-directed force has canceled with each other.

Last edited: Jun 24, 2009
14. Jun 24, 2009

### Maxwellkid

Isn't there another way to set this integral up? Can I convert all this into spherical coordinates?

I'm asking this because step by step i'm not sure how to get to above after I write down the equation dF3= IdL x B. B = <0, 0, -Bz>. IdL= <? , ?, ?>

15. Jun 24, 2009

### LowlyPion

Why would you want to do that?

Have you evaluated the integral of Sinθ from 0 to π yet?

Why try to thrash about on it?

16. Jun 24, 2009

### Maxwellkid

Yes, and I got -2.

What I'm wondering is this step:

dF3= dF3 cos@ i + dF3 sin@ j. = IR d@ x B

IR d@ =<?, ?, ?>
B = <0, 0, -Bz>

17. Jun 24, 2009

### LowlyPion

OK so you know then that the vertically directed force in the j direction is going to be proportional to 2R. (The same as if you had a straight wire 2R long bridging the same gap.)

The i components, if you were to evaluate that integral will be the integral of cosθ from 0 to π, which looks to me to be 0. I ignored their contribution, recognizing from symmetry that for each wire element with +i directed force, there is an equal and opposite -i directed force element. The sum of all those will be 0.

18. Jun 24, 2009

### Maxwellkid

I'm not sure how (IR d@ x B) distributes into dF cos@ i + dF sin@ j

(IRd@ x B) cos@ i + (IRd@ x B) sin@ j

How would you evaluate this integral?

19. Jun 24, 2009

### Maxwellkid

Thanks for pointing out the symmetry that cancels out the I component. I see that, but I don't see the technique to evaluate the j component

20. Jun 24, 2009

### LowlyPion

Note I wrote my force component equation based on the cross product already taken. I was expressing the components of the force on the individual wire elements as a function of θ from intuition. Then taking the integral.

I think the current-wire elements are originally directed as <I*sinθ, I*cosθ, 0> This gets crossed with <0, 0, B*k>

I think this results in <B*I*cosθ, B*I*sinθ, 0> which I integrated to give
<0, 2R*B*I, 0>

21. Jun 24, 2009

### Maxwellkid

I don't see the above. This is true: Sin@ = I/R. Or sin@= L/R. R= radius of semicircle

Last edited: Jun 24, 2009
22. Jun 24, 2009

### LowlyPion

Aren't those the tangential elements of the circle ... along the direction that the current is flowing?

23. Jun 24, 2009

### Maxwellkid

24. Jun 24, 2009

### Maxwellkid

Okay, I think I've got it

25. Jun 24, 2009

### Maxwellkid

Thank you Pion!!!

Do magnetic fields attract or repel charges? If I were to have a static electron next to the North pole of a magnet with no acceleration from both the charge nor the magnet, will I produce any force????