Calculate the size of the magnetic field 20 m below a high voltage power line. The line carries $\mathbf{450}\mathbf{MW}$at a voltage of $\mathbf{300}\mathbf{,}\mathbf{000}\mathbf{}\mathbf{V}$

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$\mathrm{P}=450\hspace{0.33em}\mathrm{MW}\frac{{10}^6\hspace{0.33em}\mathrm{W}}{1\hspace{0.33em}\mathrm{MW}}=4.50\times {10}^8\mathrm{W}$

The voltage across the line is$\mathrm{\Delta V}=300,000\hspace{0.33em}\mathrm{V}=3\times {10}^5\hspace{0.33em}\mathrm{V}$

Current flowing in the line = I

Strength of the magnetic field = B

Distance from the line is,$\mathrm{r}=20\hspace{0.33em}\mathrm{m}$

Current flowing is given as

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

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

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

The current passing through the wire is denoted by

$\mathrm{I}=\frac{\mathrm{p}}{\mathrm{\Delta V}}$

Substitute the values

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

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

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

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