91Ó°ÊÓ

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

Resistance per unit length of wire,$\frac{\text{R}}{\text{L}}=\frac{1}{300}\text{Ω/³¾}$

Current through the wire,$\text{I}=25\text{‼î}$

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,$âˆ\circledR Bâ‹\dots ds={\mathrm{μ}}_{0}I$(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,$\stackrel{→}{S}=\frac{\stackrel{→}{E}×\stackrel{→}{B}}{{\mathrm{μ}}_0}$(4)

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

$\begin{array}{c}E=\frac{IR}{L}\\ =25\text{ }A×\frac{1}{300}\text{ }\mathrm{Î\copyright }/m\\ =0.0833\text{ V/³¾}\end{array}$

Hence, the electric field at point P is $0.0833\text{ V/³¾}$along the x-axis.

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

$âˆ\circledR ds=\mathrm{Ï€}d$

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" $\begin{array}{c}B=\frac{{\mathrm{μ}}_{0}I}{\mathrm{Ï€}d}\\ =\frac{4\mathrm{Ï€}×{10}^{−7}\text{ }H/m×I}{\mathrm{Ï€}d}\\ =\frac{4×{10}^{−7}\text{ }H/m×25\text{ }A}{3.14×2.50×{10}^{−3}\text{ }m\text{ }}\\ =4.00×{10}^{−3}\text{â€Í¿}\left(\frac{1\text{ }mT}{{10}^{−3}\text{â€Í¿}}\right)\\ =4.00\text{″¾T}\end{array}$

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.$4.00\text{″¾T}$

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

$\begin{array}{c}\stackrel{→}{S}=\frac{0.0833\text{ }V/m×4×{10}^{−3}\text{ }T}{4\mathrm{Ï€}×{10}^{−7}\text{ }H/m}\\ =265{\text{ W/³¾}}^{\text{2}}\end{array}$

Hence, the value of the Poynting vector is.$265{\text{ W/³¾}}^{\text{2}}$

Here,$\stackrel{â‡\P Ä}{E}$ is along the x-axis and$\stackrel{â‡\P Ä}{B}$ is along the z-axis.

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

$\begin{array}{c}\stackrel{→}{S}=\frac{E\stackrel{→}{\widehat{i}}×B\stackrel{→}{\widehat{k}}}{{\mathrm{μ}}_0}\\ =\frac{I\widehat{i}×\frac{R}{L}×\frac{{\mathrm{μ}}_{0}I}{\mathrm{π}d}\widehat{k}}{{\mathrm{μ}}_0}\left(\text{fromequations(1)and(3)}\right)\\ =\frac{I^{2}R}{\mathrm{π}dL}(\widehat{i}×\widehat{k})\\ =\frac{I^{2}R}{\mathrm{π}dL}(−\widehat{j})\end{array}$

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

The negative indicates that it is directed inwards.