91Ó°ÊÓ

Write the expression for the final angular momentum in the fields.

$\mathit{L}\mathbf{=}\mathbf{∫}\mathit{l}\mathit{d}\mathit{τ}$ …… (1)

Here, l is the angular momentum density and $∫d\mathrm{τ}$is the cross-sectional area.

(a)

Write the expression for the angular momentum density.

I = r x g …… (2)

Here, g is the momentum density.

Write the expression for the momentum density.

$g={\mathrm{Î\mu }}_0\left(E×B\right)$ …… (3)

Here, E is the electric field, and B is the magnetic field produced by the solenoid at radius b.

Write the expression for the electric field.

$$\begin{gathered} E=\frac{1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{λ}}{s}\hat{s} \\ E=\frac{Q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}ls}\hat{s} \end{gathered}$$

Write the expression for the magnetic field produced by the solenoid at radius b.

$B={\mathrm{μ}}_{0}K\hat{z}$

Substitute the standard formula for K.

localid="1653482230148" $B={\mathrm{μ}}_0\left({\mathrm{σ}}_b{\mathrm{Ó\neg }}_{b}b\right)\stackrel{Áåž}{z}$

Substitute all the known values in equation (3).

localid="1653482236658" $g={\mathrm{Î\mu }}_0\left(\frac{Q}{2{\mathrm{Ï€Î\mu }}_0/\mathrm{s}}\stackrel{Áåž}{s}×\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_{b}Q}{2\mathrm{Ï€\pm õ}}\stackrel{Áåž}{z}\right)$

localid="1653482242686" $g={\mathrm{Î\mu }}_0\left(\frac{Q}{2{\mathrm{Ï€Î\mu }}_{0}I\mathrm{s}}\right)\left(-\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_{b}Q}{2\mathrm{Ï€\pm õ}}\right)\left(\stackrel{Áåž}{s}×\stackrel{Áåž}{z}\right)$

localid="1653482248181" $g=\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{\mathrm{I}}^2\mathrm{s}}\stackrel{Áåž}{âˆ\dots }$

Substitute the value of g in equation (2).

localid="1653482256169" $I=r×\left(\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{\mathrm{I}}^2\mathrm{s}}\stackrel{Áåž}{âˆ\dots }\right)$

localid="1653482262657" $I=\left(\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{\mathrm{I}}^2\mathrm{s}}\right)\left(r×\stackrel{Áåž}{âˆ\dots }\right)$

localid="1653482269792" $I=\left(\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{\mathrm{I}}^2}\right)\stackrel{Áåž}{z}$

Substitute the known values in equation (1).

$L=\left(\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{\mathrm{I}}^2}\right)\stackrel{Áåž}{Z}∫d\mathrm{Î\P }$

localid="1653482294398" $$\begin{gathered} L=\left(\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{l}^2}\right)\hat{z}∫d\mathrm{Ï„} \\ L=\left(\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2}{4{\mathrm{Ï€}}^2{l}^2}\right)\mathrm{Ï€}\left(b^2-a^2\right)l\hat{z} \\ L=\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2\left(b^2-a^2\right)}{4\mathrm{Ï€}l}\hat{z} \end{gathered}$$

Therefore, the final angular momentum in the fields is localid="1653482299337" $L=\frac{{\mathrm{μ}}_0{\mathrm{Ó\neg }}_b{Q}^2\left(b^2-a^2\right)}{4\mathrm{Ï€}l}\hat{z}$.

(b)

Write the expression for the extra electric field induced by the changing in the magnetic field due to the rotating shell.

$$\begin{gathered} \boxed{∫}E·dI=-\frac{dâˆ\dots }{dt} \\ E\left(2\mathrm{Ï€²õ}\right)=-\frac{dâˆ\dots }{dt} \\ E=-\frac{1}{2\mathrm{Ï€²õ}}\frac{dâˆ\dots }{dt} \end{gathered}$$

…… (4)

Write the expression for the electric flux .

$$\begin{gathered} âˆ\dots =\frac{{\mathrm{μ}}_{0}Q}{2\mathrm{Ï€\pm õ}}\left({\mathrm{Ó\neg }}_a-{\mathrm{Ó\neg }}_b\right){\mathrm{Ï€²¹}}^2-\frac{{\mathrm{μ}}_0{\mathrm{QÓ\neg }}_{\mathrm{b}}}{2\mathrm{Ï€\pm õ}}\mathrm{Ï€}\left({\mathrm{s}}^2-{\mathrm{a}}^2\right) \\ âˆ\dots =\frac{{\mathrm{μ}}_0\mathrm{Q}}{2\mathrm{Ï€\pm õ}}\left({\mathrm{Ó\neg }}_{\mathrm{a}}{\mathrm{a}}^2-{\mathrm{Ó\neg }}_{\mathrm{b}}{\mathrm{s}}^2\right) \end{gathered}$$

Substitute the known values in equation (4).

$$\begin{gathered} E=-\frac{1}{2\mathrm{Ï€²õ}}\frac{\left[{\displaystyle \frac{{\mathrm{μ}}_{0}Q}{2I}}\left({\mathrm{Ó\neg }}_a{a}^2-{\mathrm{Ó\neg }}_b{s}^2\right)\right]}{dt}\stackrel{Áåž}{âˆ\dots } \\ E=-\frac{1}{2\mathrm{Ï€²õ}}\frac{{\mathrm{μ}}_{0}Q}{2I}\left(a^2\frac{d{\mathrm{Ó\neg }}_a}{dt}-s^2\frac{d{\mathrm{Ó\neg }}_b}{dt}\right)\stackrel{Áåž}{âˆ\dots } \end{gathered}$$

For radius a and b, the induces electric field will be,

$E\left(a\right)=-\frac{{\mathrm{μ}}_{0}Qa}{4\mathrm{Ï€\pm õ}}\left(\frac{d{\mathrm{Ó\neg }}_a}{dt}-\frac{d{\mathrm{Ó\neg }}_b}{dt}\right)\stackrel{Áåž}{âˆ\dots }$and$E\left(b\right)=-\frac{{\mathrm{μ}}_{0}Qa}{4\mathrm{Ï€\pm õ}}\left(a^2\frac{d{\mathrm{Ó\neg }}_a}{dt}-b^2\frac{d{\mathrm{Ó\neg }}_b}{dt}\right)\stackrel{Áåž}{âˆ\dots }$

Write the expression for the torque on a shell.

$$\begin{gathered} N=r×qE \\ N=qsE\stackrel{Áåž}{z} \end{gathered}$$

For radius a:

$$\begin{gathered} N_a=Q_a\left(-\frac{{\mathrm{μ}}_{0}Qa}{4\mathrm{Ï€\pm õ}}\right)\left(\frac{d{\mathrm{Ó\neg }}_a}{dt}-\frac{d{\mathrm{Ó\neg }}_b}{dt}\right)\stackrel{Áåž}{z} \\ {L}_a={∫}_0^{∞}{N}_a=-\frac{{\mathrm{μ}}_0{Q}^2{a}^2}{4\mathrm{Ï€\pm õ}}\left({\mathrm{Ó\neg }}_a-{\mathrm{Ó\neg }}_b\right)\stackrel{Áåž}{z} \end{gathered}$$

For radius b:

$$\begin{gathered} N_b=-Q_b\left(-\frac{{\mathrm{μ}}_{0}Q}{4\mathrm{Ï€\pm õ}b}\right)\left(a^2\frac{d{\mathrm{Ó\neg }}_a}{dt}-b^2\frac{d{\mathrm{Ó\neg }}_b}{dt}\right)\stackrel{Áåž}{z} \\ {L}_b={∫}_0^{∞}{N}_b=-\frac{{\mathrm{μ}}_0{Q}^2}{4\mathrm{Ï€\pm õ}}\left(a^2{\mathrm{Ó\neg }}_a-b^2{\mathrm{Ó\neg }}_b\right)\stackrel{Áåž}{z} \end{gathered}$$

Hence, the extra angular momentum will be,

$$\begin{gathered} L_{tot}=L_a+L_B \\ {L}_{tot}=\left(-\frac{{\mathrm{μ}}_0{Q}^2{a}^2}{4\mathrm{Ï€\pm õ}}\left({\mathrm{Ó\neg }}_a-{\mathrm{Ó\neg }}_b\right)\stackrel{Áåž}{z}\right)+\left(-\frac{{\mathrm{μ}}_0{Q}^2}{4\mathrm{Ï€\pm õ}}\left(a^2{\mathrm{Ó\neg }}_a-b^2{\mathrm{Ó\neg }}_b\right)\stackrel{Áåž}{z}\right) \\ {L}_{tot}=-\frac{{\mathrm{μ}}_0{Q}^2{\mathrm{Ó\neg }}_b}{4\mathrm{Ï€\pm õ}}\left(b^2-a^2\right)\stackrel{Áåž}{z} \end{gathered}$$

So, the reduction in the final angular momentum in part (b) is equal to the residual angular momentum in part (a).

Therefore, the resulting extra angular momentum is $L_{tot}=-\frac{{\mathrm{μ}}_0{Q}^2{\mathrm{Ó\neg }}_b}{4\mathrm{Ï€\pm õ}}\left(b^2-a^2\right)\stackrel{Áåž}{z}$.