91Ó°ÊÓ

1. The current is, $i=50 \mathrm{A}$.
2. The diameter is,$d=2.6 \text{mm}$.

We use the formula of magnetic field due to a long straight wire at perpendicular distancefrom the wire to calculate the magnitude of magnetic field.

Formula:

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{Ï€»å}}$

The magnetic field due to the long straight wire at any distance from the wire is given by

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}d}$

We have to find the magnitude of the magnetic field on the surface of the wire whose distance from the center of wire is $R$and given by

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

Diameter $d=2.6 \text{mm}$and radius $R=\frac{2.6}{2} \text{mm}=0.0013 \text{m}$.

Substitute all the value in the above equation.

localid="1664177943519" $B=\frac{{\displaystyle 4\mathrm{π}×{10}^{-7}\raisebox{1ex}{$\mathrm{Tm}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right. ×50 \text{A}}}{{\displaystyle 2\mathrm{π}×0.0013 \text{m}}}$

localid="1663008809868" $B=7.7×{10}^{-3} \text{T}$

Hence the net magnetic field at the surface of wire is $7.7×{10}^{-3} \text{T}$.