Question

A cylindrical cable of radius $\mathbf{8}\mathbf{}\mathbf{ }\mathbf{\text{mm}}$carries a current of$\mathbf{25}\mathbf{}\mathbf{A}$, uniformly spread over its cross-sectional area. At what distance from the center of the wire is there a point within the wire where the magnetic field magnitude is$\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{\text{mT}}$?

Short answer

The distance from the centre of the wire where the magnetic field is $0.100\text{mT}$ is$0.00128\text{m}$.

Step-by-step solution

Identification of given data

1. Radius of cable$R=8 \text{mm}$
2. Current$i=25A$
3. Magnetic field magnitude$0.100\text{mT}$

Understanding the concept of Ampere's circuital law

According to Ampere's circuital law, the number of times the algebraic total of the currents traveling through the loop equals the line integral of the magnetic field around a closed loop.

We use Ampere's loop law to find the required distance.

Formulae:

$\mathbf{\int }\overrightarrow{\mathbf{B}}\mathbf{.}\overrightarrow{\mathbf{d}\mathbf{s}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{i}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$

Determining the distance from the center of the wire where the magnetic field is  0.100 mT

We know Ampere’s law is

$\int \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{ds}}={\mu }_0{i}_{\mathrm{enc}}$

Where magnetic field $\overrightarrow{B}$is is a small length element $\overrightarrow{ds}$of the amperian loop.

We draw the amperian loop inside the wire whose radius is$r$.
$\int Bdscos\theta ={\mu }_0{i}_{enc}$

Where localid="1662979084598" $\theta$is the angle between length element and magnetic field, and it is $0$.

Because the current is uniformly distributed, the current$i_{enc}$ encircled the loop
is proportional to the area encircled by the loop; that is

localid="1662979414953" $$\begin{gathered} i_{enc}=\frac{i_{total}{A}_{enc}}{A_{total}}B\int \overrightarrow{ds} \\ ={\mu }_0\frac{i\pi r^2}{\pi R^2} \end{gathered}$$

This is the magnetic field inside the wire.

Substituting the given values in the above magnetic field we get

$$\begin{gathered} 0.100\times {10}^{-3} \text{T}=\frac{\left(1.26\times {10}^{-6}\frac{\text{T.m}}{\text{A}}\right)\left(25\text{A}\right)r}{2\pi \times 64\times {10}^{-6}{\text{m}}^{\text{2}}} \\ r=\frac{\left(0.100\times {10}^{-3} \text{T}\right)\left(2\pi \times 64\times {10}^{-6}{\text{m}}^{\text{2}}\right)}{\left(1.26\times {10}^{-6}\frac{\text{T.m}}{\text{A}}\right)\left(25 \text{A}\right)} \end{gathered}$$

localid="1663054585587" $r=1.28\times {10}^{-3}\text{m} \text{or} 0.00128 \text{m}$

the distance from the center of the wire is$1.28\times {10}^{-3}\mathrm{m}$.