Find the emf induced in a metal ring rotating in a magnetic field

[im_stupid]

First off, sorry if this is a simple question, I'm very bad at electromagnetism.

Homework Statement

A metal ring of radius R rotates with constant angular velocity ω about a diameter. Perpendicular to the rotation axis is a constant magnetic induction field \underline{B}. Find the EMF induced in the ring as a function of time.

Homework Equations

\omega=d\phi/dt (1)
\Phi=\int\underline{B}.\underline{dA} (2)
emf=-d\Phi/dt (3)

The Attempt at a Solution

The area, through with the magnetic field acts, changes with time. Find an expression for dA/dt:
dA=rdrd\theta (4)
in polar coordinates, θ is the angle between ω and r.

The the r coordinate of the area, through with the magnetic field acts, changes with time.
r=Rcos\omega t
dr/d\phi=-Rsin\omega t
dr/dt=(dr/d\phi)(d\phi/dt)=-R\omega sin\omega t

∴ inserting into (4)
dA/dt = -\pi R^{2}\omega^{2}cos\omega tsin\omega t

So the emf would be
emf=-d\Phi/dt=BdA/dt=B\pi R^{2}\omega^{2}cos\omega tsin\omega t

Am I along the right lines? Or am I over complicating things?

 

[collinsmark]

Hi im_stupid,

Welcome to Physics Forums!

First off, sorry if this is a simple question, I'm very bad at electromagnetism.

Homework Statement

A metal ring of radius R rotates with constant angular velocity ω about a diameter. Perpendicular to the rotation axis is a constant magnetic induction field \underline{B}. Find the EMF induced in the ring as a function of time.

Homework Equations

\omega=d\phi/dt (1)
\Phi=\int\underline{B}.\underline{dA} (2)
emf=-d\Phi/dt (3)

The Attempt at a Solution

The area, through with the magnetic field acts, changes with time. Find an expression for dA/dt:
dA=rdrd\theta (4)
in polar coordinates, θ is the angle between ω and r.

The the r coordinate of the area, through with the magnetic field acts, changes with time.
r=Rcos\omega t
dr/d\phi=-Rsin\omega t
dr/dt=(dr/d\phi)(d\phi/dt)=-R\omega sin\omega t

∴ inserting into (4)
dA/dt = -\pi R^{2}\omega^{2}cos\omega tsin\omega t

So the emf would be
emf=-d\Phi/dt=BdA/dt=B\pi R^{2}\omega^{2}cos\omega tsin\omega t

Am I along the right lines? Or am I over complicating things?

Me thinks you're over-complicating things -- well, regarding the area anyway.

Consider that at t = 0, the axis of the loop is parallel to the magnetic field \vec B. (I.e. the plane of the loop is perpendicular to the direction of the field.) And let's call the area of the loop A.

The flux at this point is simply \Phi_{t=0} = AB.

When the loop rotates around in a circle (along the axis of a diameter), the amount of flux fluctuates too sinusoidally, \Phi \left( t \right) = AB \cos \omega t.
(Or more generally we can write \Phi \left( t \right) = AB \cos \left( \omega t + \phi_0 \right), if the starting position was an arbitrary angle \phi_0.)

You might object, saying, "but when the loop spins around 180^o, the area is at maximum again with respect to the field." But not really. when the loop has turned around 180^o, the area is now in the opposite direction it was with respect to the magnetic field, and the emf has done a corresponding reversal with respect to the loop's frame of reference. The emf is now negative.

--------------------

Still not convinced?

Okay, here is a little more analytical approach. Let's treat both the magnitic field and the area as vectors. Consider the magnetic field \vec B points along the x-axis.
\vec B = B \hat \imath,
and it's a constant so it doesn't change with time.

Now consider the loop, with area A (I'll let you calculate the area of a circle with radius R) rotates along the z-axis. The direction of the area vector is the same direction as the normal vector -- a vector perpendicular to the surface plane. And this vector is rotating in a circle. (Its direction is time varying.)
\vec A = A \cos \left( \omega t + \phi_0 \right) \hat \imath + A \sin \left( \omega t + \phi_0 \right) \hat \jmath
Now just take the dot product of those two vectors. Note that the dot product is a time varying scalar.
\Phi \left( t \right) = \vec B \cdot \vec A
Good luck! (You should be able to take it from here. )

 

[im_stupid]

Thanks very much, that was the jump in logic that I needed.