Along the path shown in Figure 21.46 the magnetic field is measured and is found to be uniform in magnitude and always tangent to the circular path. If the radius of the path is $\mathbf{3}\mathbf{ }\mathbf{\text{cm}}$and $\mathit{B}$along the path is $\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{\text{T}}$, what are magnitude and direction of the current enclosed by the path?

Strength of the magnetic field$B=1.3\times {10}^{-6} \text{T}$

Radius of the path$R=3 \text{cm} \text{or} \text{0.03} \text{m}$

According to Ampere's circuital equation, the line integral of the magnetic field forming a closed loop around a current-carrying wire in a plane normal to the current is equal to ${\mathit{\mu }}_{\mathbf{o}}$times the net current flowing through the loop.

$\oint \overrightarrow{B}·d\overrightarrow{L}={\mu }_{o}I$ ...(i)

Where, $B$is the strength of the magnetic field, $d\overrightarrow{L}$is the length of elementary part, ${\mu }_o$is the permeability of the medium ($4\pi \times {10}^{-7} \text{H/m}$) and$I$is the strength of current

By using equation (i)

$\begin{array}{rcl}\oint \overrightarrow{B}·d\overrightarrow{L}& =& B\oint d\overrightarrow{L}\\ & =& B\left(2\pi R\right)\end{array}$

$\begin{array}{rcl}B\left(2\pi R\right)& =& {\mu }_{o}I\\ I& =& \frac{B\times \left(2\pi R\right)}{{\mu }_o}\end{array}$

Substitute all the values in above equation

$\begin{array}{rcl}I& =& \frac{\left(1.3\times {10}^{-6} \text{T}\right)\times \left(2\pi \times \text{0.03} \text{m}\right)}{4\pi \times {10}^{-7} \text{H/m}}\\ & =& 0.195 \text{A}\end{array}$

Hence the magnitude of the current enclose by path is $0.195 \text{A}$and the direction is into the plane of the paper