Imagine a very long solenoid with radius R, n turns per unit length, and current I . Coaxial with the solenoid are two long cylindrical (nonconducting) shells of length l—one, inside the solenoid at radius a, carries a charge +Q, uniformly distributed over its surface; the other, outside the so

Question

Imagine a very long solenoid with radius R, n turns per unit length, and current I . Coaxial with the solenoid are two long cylindrical (nonconducting) shells of length l—one, inside the solenoid at radius a, carries a charge +Q, uniformly distributed over its surface; the other, outside the solenoid at radius b, carries charge −Q (see Fig. 8.7; l is supposed to be much greater than b). When the current in the solenoid is gradually reduced, the cylinders begin to rotate, as we found in Ex. 7.8. Question: Where does the angular momentum come from?[latex]^8[/latex]

Accepted Answer

It was initially stored in the fields. Before the current was switched off, there was an electric field,

[latex]E=\frac{Q}{2\pi\epsilon _0l}\frac{1}{s}\hat{s} (a\lt s\lt b), [/latex]

in the region between the cylinders, and a magnetic field,

[latex]B=\mu_0nI\hat{z} (s\lt R),[/latex]

inside the solenoid. The momentum density (Eq. 8.29) was therefore

[latex]g=\mu_0\epsilon _0S=\epsilon _0(E imes B),[/latex] (8.29)

[latex]g=-\frac{\mu_0nIQ}{2\pi ls}\hat{\phi}, [/latex]

in the region a < s < R. The z component of the angular momentum density was

[latex](r imes g)_z=-\frac{\mu_0nIQ}{2\pi l}, [/latex]

which is constant (independent of s). To get the total angular momentum in the fields, we simply multiply by the volume, [latex]π(R^2 − a^2)l:^9[/latex]

[latex]L=-\frac{1}{2}\mu_0nIQ(R^2-a^2)\hat{z}. [/latex] (8.34)

When the current is turned off, the changing magnetic field induces a circumferential electric field, given by Faraday’s law:

[latex]E=\begin{cases}-\frac{1}{2}\mu_0n\frac{dI}{dt}\frac{R^2}{s}\hat{\phi}, (s\gt R) \\-\frac{1}{2}\mu_0n\frac{dI}{dt}s\hat{\phi}, (s\lt R)\end{cases} [/latex]

Thus the torque on the outer cylinder is

[latex]N_b=r imes (-QE)=\frac{1}{2}\mu_0nQR^2\frac{dI}{dt}\hat{z}, [/latex]

and it picks up an angular momentum

[latex]L_b=\frac{1}{2}\mu_0nQR^2\hat{z}\int_{1}^{0}{\frac{dI}{dt}dt }=-\frac{1}{2}\mu_0nIQR^2\hat{z} .[/latex]

Similarly, the torque on the inner cylinder is

[latex]N_a=-\frac{1}{2}\mu_0nQa^2\frac{dI}{dt}\hat{z},[/latex]

and its angular momentum increase is

[latex]L_a=-\frac{1}{2}\mu_0nIQa^2\hat{z}.[/latex]

So it all works out: [latex]L_{em} = L_a + L_b.[/latex] The angular momentum lost by the fields is precisely equal to the angular momentum gained by the cylinders, and the total angular momentum (fields plus matter) is conserved.

Question 8.4: Imagine a very long solenoid with radius R, n turns per unit...

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