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I'm pretty new to electrodynamics, and I was wondering whether someone could help me with the following problem:

A thin conducting rod with length $\ell$ rotates at rate $\omega$ in a uniform magnetic field.

(a) If the magnetic field $B$ is constant with respect to time, what is an expression for the induced voltage between the two ends of the rods? Moreover, which end of the rod is at higher voltage (the part that's rotating, or the point that's fixed) and in which direction do the electrons flow? Assume $B$ points out of the page, and the rod is rotating counterclockwise.

(b) Now suppose a conducting ring is added. The end of the rotating rod is always in contact with it. How do the answers in the preceding part change?

(c) Repeat the first part but now assume the magnetic field is time-dependent. Say it takes the form $B(t) = B_0 \sin(\omega_0 t)$ where $\omega_0$ is much smaller than $\omega$ so that the field oscillates in and out of the page.

I think I need to use Faraday's Law of Induction for part (a), which says the induced EMF is $-d\Phi_{B}/dt$. However, since the B-field is constant with respect to time (and $\Phi_{B} = \vec{B} \cdot \vec{A})$, would this just mean that the answer to the first part of part (a) is zero? I'm not entirely sure which part has a higher voltage due to my previous answer. I think that the last part of this problem involves Lenz's law, but I'm not too sure about how to approach this either.

Can someone please help me with this exercise?

2 Answers2

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Which part of the rod would be at higher voltage depends on the direction of the field with respect to $\vec \omega$.

Consider a small length element of the rod. deduce its instantaneous velocity and use it to determine which end would be higher in potential.

To find the magnitude, 'rate of change of flux' (this might seem absurd since the rod is not closed in the sense of B field passing through it) can be interpreted as $B *dA/dt$ where $dA/dt$ is the rate of area swept by the rod.

aneet kumar
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Consider in a time $dt$ the rod moves an angle of $\omega dt$ and hence the area of sector covered is just $\pi l^2 \cdot \frac{\omega dt}{2\pi}$ thus the $$|d\phi|={B} \omega \frac{l^2}{2}dt \iff |\frac{d\phi}{dt}|=\frac{B\omega l^2}{2}$$ Now consier an electron on the rod it is easy to see using $F=q\vec{v}\times \vec{B}$

any electron will be forced inwards of rod which means pivot will be at lower potential.

For the third part we have another factor : the changing magnetic field the following formula maybe useful$$\frac{d\phi}{dt}=- \vec{B(t)}\cdot \frac{dA}{dt}-\underbrace{{\vec{A}\cdot\frac{dB(t)}{dt}}}_{=0}$$ I leave the rest to you......

Note that $\frac{dA}{dt}=\frac{\omega l^2}{2}$ as we found in first part may prove to be useful .

Hari Ramakrishnan Sudhakar
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