91Ó°ÊÓ

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

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

Here, N is the torque on the spoke.

Write the expression for the torque on the spoke.

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

Here, r is the position vector.

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

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

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

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

$B=\mathrm{μ}nl\hat{z}$

Write the expression for the length element.

$dl=dr\hat{r}$

Substitute the known values in equation (3).

$$\begin{gathered} dF=l\text{'}\left(dr\hat{r}\right)\left({\mathrm{μ}}_{0}nl\hat{z}\right) \\ dF=l\text{'}{\mathrm{μ}}_{0}nldr\left(\hat{r}×\hat{z}\right) \\ dF=l\text{'}{\mathrm{μ}}_{0}nldr\mathrm{ϕ} \end{gathered}$$

Write the expression for the position vector.

$r=r\hat{r}$

Substitute the known values in equation (2).

$$\begin{gathered} N={∫}_0^R\left(r\hat{r}\right)×\left(-l\text{'}{\mathrm{μ}}_{0}nldr\mathrm{ϕ}\right) \\ N=l\text{'}{\mathrm{μ}}_{0}nl{∫}_0^{r}rdr\left(-\hat{r}×\widehat{\mathrm{ϕ}}\right) \\ N=l\text{'}{\mathrm{μ}}_{0}nl{\left(\frac{r^2}{2}\right)}_0^R\left(\hat{z}\right) \\ N=-\left(\frac{1}{2}\right)l\text{'}{\mathrm{μ}}_{0}nl\left(R^2-a^2\right)\left(\hat{z}\right) \end{gathered}$$

Substitute the known values in equation (1).

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

Consider the expression for the charge on the cylinder.

$Q=∫l\text{'}dt$

Rewrite the equation for the angular momentum.

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

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

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

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