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$\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{\text{mT}}$?

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

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:

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

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$1.28\times {10}^{-3}\mathrm{m}$.