Square Loop Moving Through Varying Magnetic Field

[trevor51590]

Homework Statement

A square loop of side length L is moving in the x direction with velocity v. v=v_x. B points along the z axis and is defined by B=αxe_z where α is a constant. Find EMF along the loop

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

Homework Equations

ε=-d\Phi_B/dt

The Attempt at a Solution

I figured that \Phi_B=∫B\cdotda=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\cdotda" 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!

 

[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?

 

[trevor51590]

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?

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?

I appreciate your reply!

 

[rude man]

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

 

[Sunil Simha]

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

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...

 

[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!

 

[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

 

[trevor51590]

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

 

[Sunil Simha]

You are welcome