Question

An infinite cylinder of radius R carries a uniform surface charge $\mathbf{\sigma }$. We propose to set it spinning about its axis, at a final angular velocity $\mathbf{\omega }$. How much work will this take, per unit length? Do it two ways, and compare your answers:

(a) Find the magnetic field and the induced electric field (in the quasistatic approximation), inside and outside the cylinder, in terms of $\mathbf{\omega }\mathbf{,}\mathbf{\omega }\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{s}$(the distance from the axis). Calculate the torque you must exert, and from that obtain the work done per unit length$\left(W=\int \mathrm{Nd\varphi }\right)$.

(b) Use Eq. 7.35 to determine the energy stored in the resulting magnetic field.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{total}\mathrm{torque}\mathrm{exerted}\mathrm{to}\mathrm{obtain}\mathrm{the}\mathrm{work}\mathrm{done}\mathrm{per}\mathrm{unit}\mathrm{length}\mathrm{is}-\frac{{\mathrm{\mu }}_0\mathrm{\pi }}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}{\mathrm{R}}^2\right)}^2. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{energy}\mathrm{stored}\mathrm{in}\mathrm{the}\mathrm{resulting}\mathrm{magnetic}\mathrm{field}\mathrm{is}\frac{{\mathrm{\mu }}_0\mathrm{\pi I}}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}{\mathrm{R}}^2\right)}^2. \end{gathered}$$

Step-by-step solution

Given information

The radius of the infinite cylinder is, R .

The uniform surface charge on the cylinder is, $\sigma$.

The final angular velocity of the spinning cylinder is, ${\omega }_f$.

Magnetic force on a solenoid

Consider for a current carrying solenoid of certain radius is kept in an electric or magnetic field and rotated about a certain axis then it experiences a force. The force experienced by the solenoid is described as the ‘Lenz’s magnetic force’.

The value of the force acting on the solenoid relies upon the direction of the movement of the solenoid and amount of supplied current.

Step 3(a): Determine the torque exerted to obtain work done per unit length

$\left(s>R\right)$The formula for the magnetic field inside the solenoid $\left(s<R\right)$is given by,

$$\begin{gathered} B={\mu }_{0}K\hat{z} \\ B={\mu }_0\sigma \omega R\hat{z} \end{gathered}$$

And, the magnetic field outside the solenoid $\left(s>R\right)$is given by,

B=0

The formula for the induced electric field inside the solenoid $\left(s<R\right)$due to the change in the magnetic field of the solenoid is given by,

$$\begin{gathered} \mathrm{E}=-\frac{\mathrm{S}}{2}\frac{\mathrm{dB}}{\mathrm{dt}}\hat{\mathrm{\varphi }} \\ \mathrm{E}=-\frac{\mathrm{sR}}{2}{\mathrm{\mu }}_0\mathrm{\sigma }\stackrel{.}{\mathrm{\omega }}\hat{\mathrm{\varphi }} \end{gathered}$$

And, the induced electric field outside the solenoid $\left(s>R\right)$due to the change in the magnetic field of the solenoid is given by,

$$\begin{gathered} \mathrm{E}=-\frac{{\mathrm{R}}^2}{2\mathrm{s}}\frac{\mathrm{dB}}{\mathrm{dt}}\hat{\mathrm{\varphi }} \\ \mathrm{E}=-\frac{{\mathrm{R}}^3}{2\mathrm{s}}{\mathrm{\mu }}_0\mathrm{\sigma }\stackrel{.}{\mathrm{\omega }\hat{\mathrm{\varphi }}} \end{gathered}$$

Now at the surface of the solenoid (s=R) , the value of the electric field will be,

$\mathrm{E}=-\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{R}}^2\mathrm{\sigma }\stackrel{.}{\mathrm{\omega }}\hat{\mathrm{\varphi }}$

Due to this electric field, there is a torque that acts on the cylinder length.

Assume, the length of the cylinder is $I$.

Then the formula for the torque developed (N) on the length of the cylinder is given by,

$$\begin{gathered} \mathrm{N}=-\mathrm{R}\left(\mathrm{\sigma }2\mathrm{\pi RI}\right)\left(\frac{1}{2}{\mathrm{\mu }}^0{\mathrm{R}}^2\mathrm{\sigma }\stackrel{.}{\mathrm{\omega }}\right)\hat{\mathrm{z}} \\ \mathrm{N}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4\stackrel{.}{\mathrm{\omega }}/\hat{\mathrm{Z}} \end{gathered}$$

The formula for the total torque exerted to obtain the work done (W) per unit length of the cylinder is given by,

$$\begin{gathered} \mathrm{W}=\int \mathrm{Nd\varphi } \\ \mathrm{W}=\int -{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4\int \frac{\mathrm{d\omega }}{\mathrm{dt}}\mathrm{d\varphi } \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4\int \frac{\mathrm{d\omega }}{\mathrm{dt}}\mathrm{d\varphi } \end{gathered}$$

Taking,$\mathrm{d\varphi }=\mathrm{\omega dt},$

$$\begin{gathered} \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4\int \frac{\mathrm{d\omega }}{\mathrm{dt}}\mathrm{\omega dt} \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4\int \mathrm{\omega d\omega } \end{gathered}$$

Integrating the expression between limits 0 and ${\omega }_f$,

$$\begin{gathered} \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4{\int }_0^{{\mathrm{\omega }}_{\mathrm{f}}}\mathrm{\omega d\omega } \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4{\left[\frac{{\mathrm{\omega }}^2}{2}\right]}_0^{{\mathrm{\omega }}_{\mathrm{f}}} \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{\pi \mu }}_0{\mathrm{\sigma }}^2{\mathrm{R}}^4\left[\frac{{{\mathrm{\omega }}_{\mathrm{f}}}^2}{2}-0\right] \\ \frac{\mathrm{W}}{\mathrm{I}}=-\frac{{\mathrm{\mu }}_0\mathrm{\pi }}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}{\mathrm{R}}^2\right)}^2 \end{gathered}$$

Here, the negative sign indicates that that the work is done by the field.

Hence, the total torque exerted to obtain the work done per unit length is $-\frac{{\mathrm{\mu }}_0\mathrm{\pi }}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}{\mathrm{R}}^2\right)}^2$.

Step 4(b): Determine the energy stored in the resulting magnetic field

The formula for the uniform magnetic field inside the solenoid is given by,

$$\begin{gathered} \mathrm{B}={\mathrm{\mu }}_0\mathrm{K}\hat{\mathrm{z}} \\ \mathrm{B}={\mathrm{\mu }}_0{\mathrm{\sigma \omega }}_{\mathrm{f}}\mathrm{R}\hat{\mathrm{z}} \end{gathered}$$

Using conservation of energy, the formula for the energy stored $\left(U\right)$in the resulting magnetic field is given by,

$$\begin{gathered} \mathrm{U}+\mathrm{W}=0 \\ \mathrm{U}+\left\{-\frac{{\mathrm{\mu }}_0\mathrm{\pi I}}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}\mathrm{R}\right)}^2\right\}=0 \\ \mathrm{U}=\frac{{\mathrm{\mu }}_0\mathrm{\pi I}}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}\mathrm{R}\right)}^2 \end{gathered}$$

Hence, the energy stored in the resulting magnetic field is$\frac{{\mathrm{\mu }}_0\mathrm{\pi I}}{2}{\left({\mathrm{\sigma \omega }}_{\mathrm{f}}\mathrm{R}\right)}^2$ .