Question

A cylindrical conductor with a circular cross-section has a radius a and a resistivity r and carries a constant current I. (a) What are the magnitude and direction of the electric-field vector ES at a point just inside the wire at a distance a from the axis? (b) What are the magnitude and direction of the magnetic-field vector BS at the same point? (c) What are the magnitude and direction of the Poynting vector SS at the same point? (The direction of S is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length l of the conductor. (Hint: Integrate S Sover the surface of this volume.) Compare your result to the rate of generation of thermal energy in, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

Short answer

$\mathrm{E}=\left(\frac{\mathrm{pl}}{{\mathrm{\pi a}}^2}\right)\mathrm{B})\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi a}}\mathrm{C})\mathrm{S}=\frac{{\mathrm{pl}}^2}{2{\mathrm{\pi }}^2{\mathrm{a}}^3}\mathrm{D})\mathrm{P}=\frac{{\mathrm{PLl}}^2}{{\mathrm{\pi }}^2{\mathrm{a}}^2}={\mathrm{l}}^2\mathrm{R}$

Step-by-step solution

Concept of the magnitude and direction of the electric-field vector ES at a point just inside the wire at a distance a from the axis

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 $\mathrm{V}=\mathrm{lR}=\mathrm{l}\left(\frac{\mathrm{pL}}{\mathrm{A}}\right)=\mathrm{l}\left(\frac{\mathrm{pL}}{{\mathrm{\pi r}}^2}\right)$Use this expression of V i to get Therefore,

$\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{1}{\mathrm{L}}\mathrm{l}\left(\frac{\mathrm{pL}}{{\mathrm{\pi r}}^2}\right)=\left(\frac{\mathrm{pL}}{{\mathrm{\pi a}}^2}\right)$

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

Concept of the magnetic field

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

Now, we integrate for $l=2\mathrm{\pi a}$to get B as next

$$\begin{gathered} \mathrm{B}{\left[\mathrm{l}\right]}_0^{2\mathrm{\pi a}}={\mathrm{\mu }}_0\mathrm{l} \\ 2\mathrm{\pi aB}={\mathrm{\mu }}_0\mathrm{l} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi a}} \end{gathered}$$

Concept of the magnitude and direction of the Poynting vector SS at the same point

The Poynting vector of the energy rate in the electromagnetic wave in vacuum is given by equation (in the form$\mathrm{S}=\frac{1}{{\mathrm{\mu }}_0}\mathrm{E}\times \mathrm{B}$ 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

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

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

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

Its direction is inward the cylinder

Concept of the rate of flow of energy

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 $2\mathrm{\pi aL}$. Now, plug the expressions for S and A to get P,

P=SA

$$\begin{gathered} =\frac{{\mathrm{pl}}^2}{2{\mathrm{\pi }}^2{\mathrm{a}}^3}\left(2\mathrm{\pi aL}\right) \\ =\frac{{\mathrm{pLl}}^2}{{\mathrm{\pi }}^2{\mathrm{a}}^2} \end{gathered}$$

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

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

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