91Ó°ÊÓ

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

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 ${\mathit{μ}}_{\mathbf{o}}$times the net current flowing through the loop.

$âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{L}={\mathrm{μ}}_{o}I$ ...(i)

Where, $B$is the strength of the magnetic field, $d\stackrel{→}{L}$is the length of elementary part, ${\mathrm{μ}}_o$is the permeability of the medium ($4\mathrm{π}×{10}^{-7} \text{H/m}$) and$I$is the strength of current

By using equation (i)

$\begin{array}{rcl}âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{L}& =& Bâˆ\circledR d\stackrel{→}{L}\\ & =& B\left(2\mathrm{Ï€}R\right)\end{array}$

$\begin{array}{rcl}B\left(2\mathrm{π}R\right)& =& {\mathrm{μ}}_{o}I\\ I& =& \frac{B×\left(2\mathrm{π}R\right)}{{\mathrm{μ}}_o}\end{array}$

Substitute all the values in above equation

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

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