In Ex. 8.4, suppose that instead of turning off the magnetic field (by reducing I) we turn off the electric field, by connecting a weakly conducting radial spoke between the cylinders. (We’ll have to cut a slot in the solenoid, so the cylinders can still rotate freely.) From the magnetic force on the current in the spoke, determine the total angular momentum delivered to the cylinders, as they discharge (they are now rigidly connected, so they rotate together). Compare the initial angular momentum stored in the fields (Eq. 8.34). (Notice that the mechanism by which angular momentum is transferred from the fields to the cylinders is entirely different in the two cases: in Ex. 8.4 it was Faraday’s law, but here it is the Lorentz force law.)

Write the expression for the angular momentum delivered to the cylinder.

$\mathit{L}\mathbf{=}\mathbf{\int }\mathit{N}\mathit{d}\mathit{t}$ ……. (1)

Here, N is the torque on the spoke.

Write the expression for the torque on the spoke.

$\mathit{N}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{R}}\mathit{r}\mathbf{\times }\mathit{d}\mathit{F}$ ……. (2)

Here, r is the position vector.

Write the expression for the force on the segment of the spoke.

$\mathit{d}\mathit{F}\mathbf{=}\mathit{l}\mathbf{\text{'}}\mathit{d}\mathit{l}\mathbf{\times }\mathit{B}$ …… (3)

Here, is the length element and l'is the current.

Write the expression for the magnetic field inside the solenoid for $a<r<R$.

$B=\mu nl\hat{z}$

Write the expression for the length element.

$dl=dr\hat{r}$

Substitute the known values in equation (3).

$$\begin{gathered} dF=l\text{'}\left(dr\hat{r}\right)\left({\mu }_{0}nl\hat{z}\right) \\ dF=l\text{'}{\mu }_{0}nldr\left(\hat{r}\times \hat{z}\right) \\ dF=l\text{'}{\mu }_{0}nldr\varphi \end{gathered}$$

Write the expression for the position vector.

$r=r\hat{r}$

Substitute the known values in equation (2).

$$\begin{gathered} N={\int }_0^R\left(r\hat{r}\right)\times \left(-l\text{'}{\mu }_{0}nldr\varphi \right) \\ N=l\text{'}{\mu }_{0}nl{\int }_0^{r}rdr\left(-\hat{r}\times \hat{\varphi }\right) \\ N=l\text{'}{\mu }_{0}nl{\left(\frac{r^2}{2}\right)}_0^R\left(\hat{z}\right) \\ N=-\left(\frac{1}{2}\right)l\text{'}{\mu }_{0}nl\left(R^2-a^2\right)\left(\hat{z}\right) \end{gathered}$$

Substitute the known values in equation (1).

$$\begin{gathered} L=\int \left[-\left(\frac{1}{2}\right)l\text{'}{\mu }_{0}nl\left(R^2-a^2\right)\left(\hat{z}\right)\right]dt \\ L=-\frac{1}{2}l\text{'}{\mu }_{0}nl\left(R^2-a^2\right)\int l\text{'}dt\left(\hat{z}\right) \end{gathered}$$

Consider the expression for the charge on the cylinder.

$Q=\int l\text{'}dt$

Rewrite the equation for the angular momentum.

$L=-\frac{1}{2}l\text{'}{\mu }_{0}nl\left(R^2-a^2\right)Q\hat{z}$

Now, compare the initial angular momentum stored in the fields with the above expression.

$L=-\frac{1}{2}{\mu }_{0}nl\left(R^2-a^2\right)Q\hat{z}$

Therefore, the angular momentum delivered to the cylinder is $L=-\frac{1}{2}{\mu }_{0}nl\left(R^2-a^2\right)Q\hat{z}$.