Faraday's Law of Induction: Calculating Induced emf in a Moving Square Loop

[robert25pl]

Induced emf around a closed path in a time-varying magnetic field.
A magnetic field is given in the xz-plane by B=Bo*cos(pi)(x-Uot)ay Wb/M^2. Consider a rigid square loop situated in the xz-plane with its vertices at (x,0,1), (x,0,2),(x+1,0,2) and (x+1,0,1).
1.What is the expression for the emf induced around the loop in the sense defined by connecting the above points.
2.If the loop is moving with the velocity V = U_{o}a_{x} m/s instead of being stationary what is the induced emf

This is what I got for flux. Can someone check me if I’m doing this right? If it is ok then I can go to next step. I hope, the latex code comes out right. Thanks for help.

Sorry, I should be more specific. This is the exact expression:
\ B = B_{o}cos{\Pi}(x-U_{0}t)a_{y}
So:
\psi=\int_{s}B\cdot\,ds=\int_{0}^{2} \int_{0}^{1}B_{0}cos{\Pi}(x-U_{0}t)a_{y}\cdot\, dx\,dz\,a_{y}

My problem is that I'm not sure that integral limits are correct. Thanks

 

[Gamma]

Can you double check your expression for B. For the expression you have given,

when y=0 (xz plane) , B=0. Therefore flux=0 since the loop is also on the xz plane.

 

[Gamma]

I am assuming ay is the unit vector along y direction. Magnetic field B is independant of z. It is a function of only x. Therefore, integrate only with restpect to x.

\psi=-\frac{d}{dt}\int_{s}B\cdot\,ds=-\frac{d}{dt}\int_{x}^{x+1}B_{0}cos{\Pi}(x-U_{0}t) dx

Induced emf is a function of x.

 

[robert25pl]

Thanks Gamma, I got the answer.

-2B_{0}U_{o}cos{\Pi}(x-U_{0}t)

So for part 2, emf would be the same because the loop is perpendicular to B? Is that right?

 

[Gamma]

So for part 2, emf would be the same because the loop is perpendicular to B? Is that right?

Not really. Now x is a function of t.

\psi=\int_{s}B\cdot\,ds=\int_{x}^{x+1}B_{0}cos{\Pi}(x-U_{0}t) dx

First evaluate the above integral. Then differentiate wrt to t. When you differentiate the above to find the emf, don't forget x=x(t). Use the fact that dx/dt = Uo

 

[robert25pl]

First evaluate the above integral. Then differentiate wrt to t. When you differentiate the above to find the emf, don't forget x=x(t). Use the fact that dx/dt = Uo

I did integrate and differentiate and induced emf is equal to:

-2B_{0}U_{o}cos{\Pi}(x-U_{0}t)

But in second part of problem "If the loop is moving with the velocity V = U_{o}a_{x} m/s instead of being stationary what is the induced emf"

I think that emf would be 0 becuase moving loop will produce a positive current. But I don't know how to proved it.

 

[Gamma]

I have explained how to go about it. First find \psi = \psi (x).
See my post #5. Yes, you will get zero.

regards.