Question

In Fig. 33-73, a long, straight copper wire (diameter$\mathbf{2}\mathbf{.50}\mathbf{\text{\hspace{0.17em}mm}}$ and resistance$\mathbf{1}\mathbf{.00}\mathbf{\text{\hspace{0.17em}Ωper}}\mathbf{300}\mathbf{\text{\hspace{0.17em}m}}$ ) carries a uniform current of $\mathbf{25}\mathbf{\text{\hspace{0.17em}A}}$in the positive x direction. For point P on the wire’s surface, calculate the magnitudes of (a) the electricfield,$\overrightarrow{\mathbf{E}}$(b) the magnetic field , $\overrightarrow{\mathbf{B}}$and (c) the Poynting vector $\overrightarrow{\mathbf{S}}$, and (d) determine the direction of $\overrightarrow{\mathbf{S}}$.

Short answer

a) The magnitude of the electric field for point P is.$0.0833\text{\hspace{0.17em}V/m}$

b) The magnitude of the magnetic fieldfor point P is.$4.00\text{\hspace{0.17em}mT}$

c) The Poynting vector for point P is$265{\text{\hspace{0.17em}W/m}}^{\text{2}}$

d) The direction of the Poyntingvector is along thenegative y-axis.

Step-by-step solution

The given data

Diameter of wire,$d=2.50\text{mm}\left(\frac{1m}{1000mm}\right)=2.50\times {10}^{-3}\text{m}$

Resistance per unit length of wire,$\frac{\text{R}}{\text{L}}=\frac{1}{300}\text{Ω/m}$

Current through the wire,$\text{I}=25\text{\hspace{0.17em}A}$

Understanding the concept of Brewster angle

We can use the equation for the electric field in terms of the potential and the distance. We can use the equation for potential from Ohm’s law in the equation for the electric field to find its value. Using Ampere’s law, we can find the value of the magnetic field. Using the formula for the Poynting vector in terms of the electric and the magnetic field, we can find its value.

Formulae:

The magnetic field for a given area according to Ampere’s law,$\oint B\cdot ds={\mu }_{0}I$(1)

The voltage equation using Ohm’s law,$\text{V}=\text{IR}$(2)

The electric field for a given voltage,$\text{E}=\frac{\text{V}}{\text{L}}$(3)

The Poynting vector due to the directional flux,$\overrightarrow{S}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{\mu }_0}$(4)

a) Calculation of the magnitude of the electric field 

Substituting equation (2) in equation (3) and using the given data, we can get the magnitude of the electric field as follows:

$\begin{array}{c}E=\frac{IR}{L}\\ =25A\times \frac{1}{300}\Omega /m\\ =0.0833\text{\hspace{0.17em}V/m}\end{array}$

Hence, the electric field at point P is $0.0833\text{\hspace{0.17em}V/m}$along the x-axis.

b) Calculation of the magnitude of the magnetic field 

Here, the point P is on the circumference of the wire with diameter d.

$\oint ds=\pi d$

where$\text{ds}$,is the element which is the circumference of the Amperian loop.

Thus, the magnitude of the magnetic field using the above value and the given data in equation (1) is as follows:

localid="1664201670921" $\begin{array}{c}B=\frac{{\mu }_{0}I}{\pi d}\\ =\frac{4\pi \times {10}^{-7}H/m\times I}{\pi d}\\ =\frac{4\times {10}^{-7}H/m\times 25A}{3.14\times 2.50\times {10}^{-3}m}\\ =4.00\times {10}^{-3}\text{\hspace{0.17em}T}\left(\frac{1mT}{{10}^{-3}\text{\hspace{0.17em}T}}\right)\\ =4.00\text{\hspace{0.17em}mT}\end{array}$

By using the right-hand rule, we can say that the magnetic field is in the clockwise direction when viewed along the x-axis.

Thus, at point P, the direction of the magnetic field is out of the page that is directed along z-axis and unit vector k.

Hence, the magnitude of the magnetic field at point P is.$4.00\text{\hspace{0.17em}mT}$

c) Calculation of the Poynting vector for point P 

Using the given data in equation (4), we can get the Poynting vector at the point P is given as follows:

$\begin{array}{c}\overrightarrow{S}=\frac{0.0833V/m\times 4\times {10}^{-3}T}{4\pi \times {10}^{-7}H/m}\\ =265{\text{\hspace{0.17em}W/m}}^{\text{2}}\end{array}$

Hence, the value of the Poynting vector is.$265{\text{\hspace{0.17em}W/m}}^{\text{2}}$

d) Calculation of the direction of the Poynting vector 

Here,$\stackrel{\rightharpoonup }{E}$ is along the x-axis and$\stackrel{\rightharpoonup }{B}$ is along the z-axis.

Thus, using the given data in equation (4), we can get the direction of the Poynting vector as follows:

$\begin{array}{c}\overrightarrow{S}=\frac{E\overrightarrow{\widehat{i}}\times B\overrightarrow{\widehat{k}}}{{\mu }_0}\\ =\frac{I\widehat{i}\times \frac{R}{L}\times \frac{{\mu }_{0}I}{\pi d}\widehat{k}}{{\mu }_0}\left(\text{fromequations(1)and(3)}\right)\\ =\frac{I^{2}R}{\pi dL}(\widehat{i}\times \widehat{k})\\ =\frac{I^{2}R}{\pi dL}(-\widehat{j})\end{array}$

Hence, the direction of the Poynting vector is along the negative y-axis.

The negative indicates that it is directed inwards.