Work done by rotating a ring in a magnetic field.

[Kloiper]

Homework Statement

Our teacher isn't very descriptive:
A ring of radius "a" and resistance "R" is placed at the center of a long solenoid with "n" turns (assume the solenoid is longer and wider than the ring) with its axis lined up with that of the solenoid. Find the amount of work done to rotate the ring so its axis is perpendicular to the solenoid's.

Homework Equations

For a solenoid, \vec{B}= μ_0nI
\phi = \int{\vec{B}\bullet da}
V=-\frac{d\phi}{dt}

The Attempt at a Solution

Unfortunately, he didn't say anything about a current, voltage, E or B field. (or H or D fields for that matter). So my first reaction was to make up a current "I" flowing through the solenoid.
This would cause a magnetic field \vec{B}=μ_0nI, and thus a flux of \phi=μ_0nI\pi a^2.
From there, I used \frac{dW}{dt} =-VI=\frac{d\phi}{dt}I, meaning dW=d\phi I, but I'm not sure which I that would be describing.
If it's I in the solenoid, which I'm pretty sure it's not, the answer works out fine because it is a constant and multiplies in just fine.
If it's I induced in the ring, we get dW=d\phi \frac{V}{R}=-\frac{d\phi^2}{dt}\frac{1}{R}. Since we know the change in flux will be from \phi=μ_0nI\pi a^2 to \phi=0, we know d\phi. But regardless of that, work never depends on time. Why would my term for work have some time variable in it?

I also thought that magnetic fields did no work, which means the work would be 0, which confuses me as to why our professor would ask us about something that doesn't happen.

 

[TSny]

Hello Kloiper. Welcome to PF!

\phi=μ_0nI\pi a^2.

This is the flux only when the axis of the ring is aligned with B. Can you write a general equation for $\phi$ as a function of the magnetic field $B$, the area of the ring $A$, and the angle $\theta$ between the axis of the ring and $B$? (To keep things from getting too cluttered, leave the expression in terms of $B$ and $A$. You can substitute specific expressions for these quantities at the end of the derivation.)

From there, I used \frac{dW}{dt} =-VI=\frac{d\phi}{dt}I, meaning dW=d\phi I, but I'm not sure which I that would be describing.
If it's I in the solenoid, which I'm pretty sure it's not, the answer works out fine because it is a constant and multiplies in just fine.
If it's I induced in the ring...

The rate at which electrical energy is generated in the ring is $VI_{\rm ring}$. You can use Ohm's law to express this in terms of just $V$ and $R$.

... work never depends on time. Why would my term for work have some time variable in it?

I think you'll find that the work done depends on the rate at which you rotate the ring. So, time is a factor. Since the problem does not specify the manner in which the ring is rotated, you might simplify by assuming the ring is rotated at a constant angular speed $\omega$ and express your answer in terms of $\omega$.

I also thought that magnetic fields did no work, which means the work would be 0, which confuses me as to why our professor would ask us about something that doesn't happen.

The work required to generate the electrical energy is not done by the magnetic field. You should think about the agent doing the work.

 

[Kloiper]

Thank you! I ended up getting an answer.

I left flux as d\phi=\vec{B}Acos(\theta)d\theta where A is simply the area of the ring. Then, using an assumed angular velocity, I was able to cancel a bunch of stuff out that simplified the problem to \frac{dW}{dt}=\frac{V^2}{R}, V=-\frac{d\phi}{dt} like you suggested.

 

[TSny]

Good. Note that if $\theta$ is the angle between the axis of the ring and the B field, then $\phi = BA\cos\theta$. So the differential $d\phi$ will have $\sin\theta$ rather than $\cos\theta$. But I think you'll still get the same result.