A line charge λ is glued onto the rim of a wheel of radius b, which is then suspended horizontally, as shown in Fig. 7.26, so that it is free to rotate (the spokes are made of some nonconducting material—wood, maybe). In the central region, out to radius a, there is a uniform magnetic field B0,

Question

A line charge λ is glued onto the rim of a wheel of radius b, which is then suspended horizontally, as shown in Fig. 7.26, so that it is free to rotate (the spokes are made of some nonconducting material—wood, maybe). In the central region, out to radius a, there is a uniform magnetic field [latex]\pmb{B}_{0}[/latex], pointing up. Now someone turns the field off. What happens?

Accepted Answer

The changing magnetic field will induce an electric field, curling around the axis of the wheel. This electric field exerts a force on the charges at the rim, and the wheel starts to turn. According to Lenz’s law, it will rotate in such a direction that its field tends to restore the upward flux. The motion, then, is counterclockwise, as viewed from above.
Faraday’s law, applied to the loop at radius b, says

[latex]\oint{E.dI}=E(2\pi b)=-\frac{d\Phi }{dt} =-\pi a^{2}\frac{dB}{dt}, or E=-\frac{a^{2}}{2b}\frac{dB}{dt} \hat{\pmb{\Phi }}. [/latex]

The torque on a segment of length dl is [latex] (r × F), or bλE dl.[/latex] The total torque on the wheel is therefore

[latex]N=b\lambda \left(-\frac{a^{2}}{2b}\frac{dB}{dt} \right)\oint{dl} =-b\lambda \pi a^{2}\frac{dB}{dt} ,[/latex]

and the angular momentum imparted to the wheel is

[latex]\int{Ndt}=-\lambda \pi a^{2}b \int_{B_0}^{0}{dB}=\lambda \pi a^{2}bB_{0} [/latex]

It doesn’t matter how quickly or slowly you turn off the field; the resulting angular velocity of the wheel is the same regardless. (If you find yourself wondering where the angular momentum came from, you’re getting ahead of the story!Wait for the next chapter.)
Note that it’s the electric field that did the rotating. To convince you of this, I deliberately set things up so that the magnetic field is zero at the location of the charge. The experimenter may tell you she never put in any electric field—all she did was switch off the magnetic field. But when she did that, an electric field automatically appeared, and it’s this electric field that turned the wheel.

Question 7.8: A line charge λ is glued onto the rim of a wheel of radius b...

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