91Ó°ÊÓ

A magnetic field is defined as a position in space near a magnet or an electric current where a physical field is formed by a moving electric charge applying force on another moving electric charge.

Power carried by the line$\mathrm{P}=450 \mathrm{MW}\frac{{10}^6 \mathrm{W}}{1 \mathrm{MW}}=4.50×{10}^8\mathrm{W}$

The voltage across the line is$\mathrm{Δ³Õ}=300,000 \mathrm{V}=3×{10}^5 \mathrm{V}$

Current flowing in the line = I

Strength of the magnetic field = B

Distance from the line is,$\mathrm{r}=20 \mathrm{m}$

Current flowing is given as

role="math" localid="1656411414808" $\mathrm{I}=\frac{\mathrm{p}}{\mathrm{Δ³Õ}}$

The magnetic field by the current-carrying wire is given as

role="math" localid="1656411006475" $\mathrm{B}=\left(\frac{\mathrm{μ}}{4\mathrm{π}}\right)\left(\frac{2\mathrm{I}}{\mathrm{r}}\right)$

The current passing through the wire is denoted by

$\mathrm{I}=\frac{\mathrm{p}}{\mathrm{Δ³Õ}}$

Substitute the values

$$\begin{gathered} \mathrm{I}=\frac{4.50×{10}^8\mathrm{W}}{3×{10}^5\mathrm{V}} \\ =1500\mathrm{A} \end{gathered}$$

The magnetic field by the current-carrying wire is given as

$$\begin{gathered} \mathrm{B}=\left(\frac{\mathrm{μ}}{4\mathrm{π}}\right)\left(\frac{2\mathrm{I}}{\mathrm{r}}\right) \\ \mathrm{B}=\left({10}^{-7}\mathrm{T}-\mathrm{m}/\mathrm{A}\right)\left(\frac{2×1500\mathrm{A}}{20\mathrm{m}}\right) \\ =1.5×{10}^{-5}\mathrm{T} \end{gathered}$$

Thus, the magnetic field comes out to be $1.5×{10}^{-5}\mathrm{T}$.