1. Jun 5, 2013

### assaftolko

A magnetic field of B=0.5B0z^ is present when a loop with radius r is rotated around the y axis with angular velocity of w=9 rad/s. What is the induced emf?

Well: flux = SB*dA = SBdAcosq(t) = Bcosq(t)SdA = 0.5B0*pi*r^2cosq(t)

d(flux)/dt = 0.5B0*pi*r^2*-sinq(t)*dq(t)/dt = 0.5B0*pi*r^2*-sinq(t)*9.

But I'm still stuck with the angle q which isn't given to me... What should I do?

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2. Jun 5, 2013

### Saitama

The attachment shows $B=(1/2)B_0 t^2$ and you use $B=(1/2)B_0$.

And what is S?

3. Jun 5, 2013

### barryj

I think you are mostly correct. However, what should the argument of cosq(t) be? could it be cosq(9t)?

4. Jun 5, 2013

### assaftolko

The first magnetic field is from another part of the question - if you look at the bottom you'll see the field I wrote here. S in the sign of integration ∫, forgot about the operators here :P

5. Jun 5, 2013

### assaftolko

You're right! It's analoug to x=vt for constant speed movement... thanks!

6. Jun 5, 2013

### Saitama

Ah ok but I don't think you need to deal with integrals here.

You know $\phi=\vec{B}.\vec{A}$. The direction of A is perpendicular to plane of ring. When the area vector rotates by an angle $\theta$, evaluate $\phi$.

PS: Can you post the answer so that I can check my working?