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.