# Homework Help: Square Loop Moving Through Varying Magnetic Field

1. Mar 27, 2013

### trevor51590

1. The problem statement, all variables and given/known data
A square loop of side length L is moving in the x direction with velocity v. v=vx. B points along the z axis and is defined by B=αxez where α is a constant. Find EMF along the loop

Hint: Find forces on charges in rod by virtue of motion of the rod.

2. Relevant equations
ε=-d$\Phi$B/dt

3. The attempt at a solution
I figured that $\Phi$B=∫B$\cdot$da=BL2cosθ.

Because the field is perpendicular, $\Phi$B=BL2
^^^I was told this was incorrect

So therefore ε=d(αxeL2)/dt, giving ε=αeL2v

This is incorrect I was told, and looking now I see that the B field varies with x. I was given a hint to "do the integral $\Phi$=∫B$\cdot$da" When I look at this, I think that da is constant at L2 so I don't see how this helps me. The hint also doesn't help me either. My work is attached below. Thank you!

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2. Mar 27, 2013

### Sunil Simha

Hello trevor51590,

If You imagine the area divided into infinitesimal segments along the y axis (why the Y-axis? think about it) and with each having a length L, what would be the breadth? Hence, what would be da?

Once you have figured that out, what would be the flux through one of these segments? And hence, what would be the net flux?

3. Mar 27, 2013

### trevor51590

If you divide a square into infinitesimal segments along the y axis (I am assuming because the field changes with x?), the breadth would be x (L), and I would assume area would be Ldy

This would leave

$\phi$B=αexL∫$^{y}_{0}$dy

Is this along the right track?

4. Mar 27, 2013

### rude man

never mind I'll leave this to Sunil and will follow the thread.

Last edited: Mar 27, 2013
5. Mar 27, 2013

### Sunil Simha

The length L is along the Y axis. So the breadth would be along....
So instead of dy, the breadth of the infinitesimal segment would be....

6. Mar 28, 2013

### trevor51590

Fairly certain I found out the answer to be αL2v

This is the same as I found before, but now I am using the proper method. Thanks!

7. Mar 28, 2013

### Sunil Simha

Yes the final answer that you obtain is correct. But your integral to obtain the total flux is a bit wrong.

You see,
$d\phi=axLdx$
and this needs to be integrated over the limits x and x+L (as you are not given the position of the loop) and the integral turns out to be $\phi=aL^2(\frac{L + 2x}{2})$

Though in the end you'll be just differentiating it w.r.t. time and getting the flux

8. Mar 28, 2013

### trevor51590

Absolutely, I see this from working through with your hints. Thank you for your assistance

9. Mar 28, 2013

### Sunil Simha

You are welcome