Question

A$\mathbf{10}\mathbf{-}\mathbf{gauge}$bare copper wire ( $\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{ }\mathbf{\text{mm}}$in diameter) can carry a current of $\mathbf{50}\mathbf{}\mathbf{A}$without overheating. For this current, what is the magnitude of the magnetic field at the surface of the wire?

Short answer

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

Step-by-step solution

Identification of the given data

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

Understanding the concept

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:

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{\pi d}}$

Calculate the magnitude of the net magnetic field at the surface of the wire

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

$B=\frac{{\mu }_{0}i}{2\pi d}$

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

$B=\frac{{\mu }_{0}i}{2\pi R}$

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

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

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