91影视

The radius of the infinite cylinder is, R .

The uniform surface charge on the cylinder is, $蟽$.

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

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 鈥楲enz鈥檚 magnetic force鈥�.

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

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

$$\begin{gathered} B={渭}_{0}K\hat{z} \\ B={渭}_0蟽蝇R\hat{z} \end{gathered}$$

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,

$$\begin{gathered} \mathrm{E}=-\frac{\mathrm{S}}{2}\frac{\mathrm{dB}}{\mathrm{dt}}\widehat{\mathrm{蠒}} \\ \mathrm{E}=-\frac{\mathrm{sR}}{2}{\mathrm{渭}}_0\mathrm{蟽}\stackrel{.}{\mathrm{蝇}}\widehat{\mathrm{蠒}} \end{gathered}$$

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,

$$\begin{gathered} \mathrm{E}=-\frac{{\mathrm{R}}^2}{2\mathrm{s}}\frac{\mathrm{dB}}{\mathrm{dt}}\widehat{\mathrm{蠒}} \\ \mathrm{E}=-\frac{{\mathrm{R}}^3}{2\mathrm{s}}{\mathrm{渭}}_0\mathrm{蟽}\stackrel{.}{\mathrm{蝇}\widehat{\mathrm{蠒}}} \end{gathered}$$

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

$\mathrm{E}=-\frac{1}{2}{\mathrm{渭}}_0{\mathrm{R}}^2\mathrm{蟽}\stackrel{.}{\mathrm{蝇}}\widehat{\mathrm{蠒}}$

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,

$$\begin{gathered} \mathrm{N}=-\mathrm{R}\left(\mathrm{蟽}2\mathrm{蟺搁滨}\right)\left(\frac{1}{2}{\mathrm{渭}}^0{\mathrm{R}}^2\mathrm{蟽}\stackrel{.}{\mathrm{蝇}}\right)\hat{\mathrm{z}} \\ \mathrm{N}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4\stackrel{.}{\mathrm{蝇}}/\hat{\mathrm{Z}} \end{gathered}$$

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

$$\begin{gathered} \mathrm{W}=鈭�\mathrm{狈诲蠒} \\ \mathrm{W}=鈭�-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4鈭�\frac{\mathrm{d蝇}}{\mathrm{dt}}\mathrm{诲蠒} \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4鈭�\frac{\mathrm{d蝇}}{\mathrm{dt}}\mathrm{诲蠒} \end{gathered}$$

Taking,$\mathrm{诲蠒}=\mathrm{蝇dt},$

$$\begin{gathered} \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4鈭�\frac{\mathrm{d蝇}}{\mathrm{dt}}\mathrm{蝇dt} \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4鈭�\mathrm{蝇d蝇} \end{gathered}$$

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

$$\begin{gathered} \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4{鈭�}_0^{{\mathrm{蝇}}_{\mathrm{f}}}\mathrm{蝇d蝇} \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4{\left[\frac{{\mathrm{蝇}}^2}{2}\right]}_0^{{\mathrm{蝇}}_{\mathrm{f}}} \\ \frac{\mathrm{W}}{\mathrm{I}}=-{\mathrm{蟺渭}}_0{\mathrm{蟽}}^2{\mathrm{R}}^4\left[\frac{{{\mathrm{蝇}}_{\mathrm{f}}}^2}{2}-0\right] \\ \frac{\mathrm{W}}{\mathrm{I}}=-\frac{{\mathrm{渭}}_0\mathrm{蟺}}{2}{\left({\mathrm{蟽蝇}}_{\mathrm{f}}{\mathrm{R}}^2\right)}^2 \end{gathered}$$

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 $-\frac{{\mathrm{渭}}_0\mathrm{蟺}}{2}{\left({\mathrm{蟽蝇}}_{\mathrm{f}}{\mathrm{R}}^2\right)}^2$.

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

$$\begin{gathered} \mathrm{B}={\mathrm{渭}}_0\mathrm{K}\hat{\mathrm{z}} \\ \mathrm{B}={\mathrm{渭}}_0{\mathrm{蟽蝇}}_{\mathrm{f}}\mathrm{R}\hat{\mathrm{z}} \end{gathered}$$

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

$$\begin{gathered} \mathrm{U}+\mathrm{W}=0 \\ \mathrm{U}+\left\{-\frac{{\mathrm{渭}}_0\mathrm{蟺滨}}{2}{\left({\mathrm{蟽蝇}}_{\mathrm{f}}\mathrm{R}\right)}^2\right\}=0 \\ \mathrm{U}=\frac{{\mathrm{渭}}_0\mathrm{蟺滨}}{2}{\left({\mathrm{蟽蝇}}_{\mathrm{f}}\mathrm{R}\right)}^2 \end{gathered}$$

Hence, the energy stored in the resulting magnetic field is$\frac{{\mathrm{渭}}_0\mathrm{蟺滨}}{2}{\left({\mathrm{蟽蝇}}_{\mathrm{f}}\mathrm{R}\right)}^2$ .