Question

The magnitude of the magnetic field $\mathbf{88}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$from the axis of a long straight wire is$\mathbf{7}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\mu T}$. What is the current in the wire?

Short answer

The current in the wire is $32.1\mathrm{A}$.

Step-by-step solution

Given Data

1. Distance $R=88\mathrm{cm}=0.88\mathrm{m}$
2. The magnitude of magnetic field $B=7.30\mathrm{\mu T}$

Understanding the concept

We use the formula of magnetic field due to a long straight wire at perpendicular distance from the axis to find the current in the wire.

Formula:

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

Calculate the current in the wire

The magnitude of magnetic field due to a long straight wire at perpendicular distance from axis is given by

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

Where$R$is the distance from the axis of the wire and${\mu }_o$is the permeability of the free space,$i$is the current flowing through the wire.

Solving for the current we get

$i=\frac{2\pi RB}{{\mu }_0}$

localid="1663019175921" $R=88cm=0.88m,B=7.30\mu T=7.30\times {10}^{-6}T,{\mu }_0=1.26\times {10}^{-6}T.m/A$

$i=\frac{\left(2\pi \right)\left(0.88m\right)\left(7.30\times {10}^{-6}T\right)}{4\pi \times {10}^{-7}T.m/A}$

$i=32.1A$

the current in the wire is$32.1A$.