Question

Figure 33-74 shows a cylindrical resistor oflength$I$,radius $a$, and resistivity$r$ carrying current.$i$(a) Show that the Poyntingvector at the surface of the resistor is everywhere directed normal to the surface, as shown. (b) Show that the rate at which energy flows into the resistor through its cylindrical surface, calculated by integrating thePoynting vector over this surface, is equal to the rate at which thermal energy is produced: localid="1664201793898" $\mathbf{\int }\overrightarrow{\mathbf{S}}\mathbf{\cdot }\overrightarrow{\mathbf{d}\mathbf{A}}\mathbf{=}{\mathit{i}}^{\mathbf{2}}\mathit{R}$ wherelocalid="1664201800300" $\overrightarrow{\mathit{d}\mathit{A}}$ is an element of the area on the cylindrical surface and localid="1664201803478" $\mathit{R}$ is the resistance.

Short answer

(a) Poynting vector $\overrightarrow{S}$ at the surface of the resistor is everywhere directed normal to the surface as shown.

(b) The rate at which energy flows into the resistor through its cylindrical surface, calculated by integratingthe Poynting vector over this surface, is equal to the rate at which thermal energy is produced, that is .$\int \overrightarrow{S}.\overrightarrow{dA}=i^{2}R$

Step-by-step solution

The given data: 

• The length of the cylinder is.$I$
• The radius of the cylinder is.$a$
• The current through the cylinder is.$i$
• The resistivity is.$\rho$

Understanding the concept of electric field:

By using the concept of electric potential and field, Ampere’s law, Ohm’s law, and Poynting vector we can prove parts a and b.

Formulae:

The magnetic field for a given area according to Ampere’s law,

$\oint B\cdot ds={\mu }_{0}I$ ….. (i)

Here, $B$ is the magnetic field, $ds$ is the displacement,${\mu }_0$ is the permeability of free space, and $I$ is the current.

The voltage equation using Ohm’s law,

$V=IR$ ….. (ii)

Here, $V$ is the voltage and$R$ is the resistor.

The electric field for a given voltage,

$E=\frac{V}{L}$ ….. (iii)

The Poynting vector due to the directional flux,

$\overrightarrow{S}=\frac{E\times B}{{\mu }_0}$….. (iv)

(a) Calculation of the direction of the Poynting vector:

Let us choose the unit vectors $\widehat{\theta ,}\widehat{a,}-\widehat{j}$along induced magnetic field $B$, radius $a$, current $i$respectively.

There is an electric potential difference across the cylinder.

Using equations (ii) and (iii), the electric field across the cylinder can be given as:

$EL=iR$

$\overrightarrow{E}=\frac{iR}{L}(-\widehat{j})$ ….. (v)

Here, the point P is on the circumference of the cylinder with radius $a$.

$\oint ds=2\pi a$

Where$ds$,is the element which is the circumference of the Amperian loop.

Thus, the magnetic field using equation (v) is given as follows:

$\overrightarrow{B}=\frac{{\mu }_{0}i}{2\pi a}\widehat{\theta }$ ….. (vi)

Using equations(v) and (vi) in equation (iv), you can get the Poynting vector as follows

$\begin{array}{c}\overrightarrow{S}=\frac{\frac{iR}{L}(-\widehat{j})\times \frac{{\mu }_{0}i}{2\pi a}\widehat{\theta }}{{\mu }_0}\\ =\frac{\frac{iR}{l}\times \frac{{\mu }_{0}i}{2\pi a}}{{\mu }_0}\end{array}$

$\overrightarrow{S}=\frac{iR}{l}\frac{i}{2\pi a}\left(\widehat{a}\right)$ ….. (vii)

Which shows that the vector is directed along the radius of the cylinder, and the negative sign implies that it is directed inwards.

Hence, Poynting vector $\overrightarrow{S}$at the surface of the resistor is everywhere directed normal to the surface as shown.

(b) Calculation of the rate at which energy flows into the resistor:

To prove part b, let usintegrate Poynting vector over the cylindrical surface which is nothing but rate flow of energy i.e. power of the resistor is given as follows:

$P=\oint \overrightarrow{S}dA$

$P=\overrightarrow{S}\oint dA$ ….. (viii)

The integral of surface area is equal to the circumference of the resistor multiply with the length of the resistor.

The circumference of the resistor with radius $a$ is$2\pi a$and the length of the resistor is $I$.

Thus, the area through which the energy flows can be given as:

$\oint dA=2\pi al$

Now, using the above value in equation (viii), you can get the rate of the energy flux is given as:

$\oint \overrightarrow{S}dA=\overrightarrow{S}\times 2\pi al$

Using equation (vii) in the above equation is given as follows:

$\begin{array}{c}\oint \overrightarrow{S}dA=\frac{iR}{l}\frac{i}{2\pi a}\times 2\pi al\\ =i^{2}R\end{array}$

Hence,the rate at which the energy flows into the resistor through its cylindrical surface is equal to the rate at which the thermal energy is produced.