Induced EMF in Rotating Magnetic Field with B=0.5B0z^

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 2K views
assaftolko
Messages
171
Reaction score
0
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?
 

Attachments

  • Clipboard01.jpg
    Clipboard01.jpg
    30 KB · Views: 446
Physics news on Phys.org
assaftolko said:
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?

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

And what is S?
 
I think you are mostly correct. However, what should the argument of cosq(t) be? could it be cosq(9t)?
 
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
 
barryj said:
I think you are mostly correct. However, what should the argument of cosq(t) be? could it be cosq(9t)?

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

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?