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

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A magnetic field of B=0.5B0z^ induces an emf in a rotating loop of radius r at an angular velocity of w=9 rad/s. The magnetic flux through the loop is calculated as 0.5B0*pi*r^2*cos(q(t)), where q(t) is the angle of rotation. The rate of change of flux leads to an induced emf expression involving the sine of the angle, specifically -0.5B0*pi*r^2*sin(q(t))*9. Participants discuss the need to determine the angle q and suggest using cos(q(9t)) to account for the rotation. The conversation emphasizes the relationship between magnetic flux and the area vector's orientation in the rotating system.
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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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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?
 
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