91Ó°ÊÓ

The radius of the cylinder is r =a with resistivity p and current I. The electric field E inside the cylinder is related to the voltage by $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}}$Where L is the length of the cylinder. The voltage is related to the resistance as given in Ohm's law inside the cylinder by $\mathrm{V}=\mathrm{lR}=\mathrm{l}\left(\frac{\mathrm{pL}}{\mathrm{A}}\right)=\mathrm{l}\left(\frac{\mathrm{pL}}{{\mathrm{Ï€°ù}}^2}\right)$Use this expression of V i to get Therefore,

$\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{1}{\mathrm{L}}\mathrm{l}\left(\frac{\mathrm{pL}}{{\mathrm{Ï€°ù}}^2}\right)=\left(\frac{\mathrm{pL}}{{\mathrm{Ï€²¹}}^2}\right)$

Thus, the electric field is parallel to the axis of the cylinder

To find the magnetic field, use Ampere's law to integrate over the circumference of the cylinder $∫B.dl={\mathrm{μ}}_{0}I$

Now, we integrate for $l=2\mathrm{Ï€²¹}$to get B as next

$$\begin{gathered} \mathrm{B}{\left[\mathrm{l}\right]}_0^{2\mathrm{Ï€²¹}}={\mathrm{μ}}_0\mathrm{l} \\ 2\mathrm{Ï€²¹B}={\mathrm{μ}}_0\mathrm{l} \\ \mathrm{B}=\frac{{\mathrm{μ}}_0\mathrm{l}}{2\mathrm{Ï€²¹}} \end{gathered}$$

The Poynting vector of the energy rate in the electromagnetic wave in vacuum is given by equation (in the form$\mathrm{S}=\frac{1}{{\mathrm{μ}}_0}\mathrm{E}×\mathrm{B}$ This equation shows the vector product between the electric field and the magnetic field. The electric field and the magnetic field are perpendicular to each other and to the direction of the propagation. Now, we use the expressions to find S

$$\begin{gathered} \mathrm{S}=\frac{1}{{\mathrm{μ}}_0}\mathrm{E}×\mathrm{B} \\ =\frac{1}{{\mathrm{μ}}_0}\mathrm{EBSin}90° \\ \mathrm{S}=\frac{1}{{\mathrm{μ}}_0}\mathrm{EB} \end{gathered}$$

Plug the expressions of E and B in above equation we get,

$$\begin{gathered} \mathrm{S}=\frac{1}{{\mathrm{μ}}_0}\mathrm{EB} \\ =\frac{1}{{\mathrm{μ}}_0}\left(\frac{\mathrm{pL}}{{\mathrm{Ï€²¹}}^2}\right)\frac{{\mathrm{μ}}_0\mathrm{L}}{2\mathrm{Ï€²¹}} \\ =\frac{{\mathrm{pl}}^2}{2{\mathrm{Ï€}}^2{\mathrm{a}}^3} \end{gathered}$$

Its direction is inward the cylinder

The rate of flow of energy is the power P and it is related to the Poynting of the energy and the area by Where A is the area of the cylinder and equals $2\mathrm{Ï€²¹L}$. Now, plug the expressions for S and A to get P,

P=SA

$$\begin{gathered} =\frac{{\mathrm{pl}}^2}{2{\mathrm{Ï€}}^2{\mathrm{a}}^3}\left(2\mathrm{Ï€²¹L}\right) \\ =\frac{{\mathrm{pLl}}^2}{{\mathrm{Ï€}}^2{\mathrm{a}}^2} \end{gathered}$$

From part (a), the resistance is given by $\mathrm{R}=\frac{\mathrm{pL}}{{\mathrm{Ï€²¹}}^2}$From our result. we can get the next expression for the power.

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{pLl}}^2}{{\mathrm{Ï€}}^2{\mathrm{a}}^2} \\ =\frac{\mathrm{pL}}{{\mathrm{Ï€²¹}}^2}{\mathrm{l}}^2 \\ ={\mathrm{l}}^2\mathrm{R} \end{gathered}$$

Therefore, the expression for power is${\mathrm{l}}^2\mathrm{R}$