91Ó°ÊÓ

As the loop and the wire lies in the same plane, the magnitude of the magnetic field due to both wire and loop will get added to the resultant at the center of curvature. The loop is perpendicular, and the wire lies in the plane, the components will be added to find the resultant between them.

Formula:

$B_{wire}=\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}R}$

$B_{loop}=\frac{{\mathrm{μ}}_{0}I}{2R}$

Consider the required equation is obtained as:

$B_C=B_{wire}+B_{loop}$

$B_C=\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}R}+\frac{{\mathrm{μ}}_{0}I}{2R}$

$B_C=\frac{{\mathrm{μ}}_{0}I}{2R}\left(\frac{1}{\mathrm{π}}+1\right)$

$B_C=\frac{{\mathrm{μ}}_{0}I\left(1+\mathrm{π}\right)}{2R\mathrm{π}}$

Substitute the values and solve as:

$B_C=\frac{4\mathrm{π}×{10}^{-7}×5.78×{10}^{-3}\left(1+\mathrm{π}\right)}{2×\mathrm{π}×0.0189}$

$B_C=2.53×{10}^{-7}\text{T}$

Using the right hand rule direction for the$B_C$will be in positive$\hat{k}$. So,

$B_C=\left(2.53×{10}^{-7}T\right)\hat{k}$

Now the loop is perpendicular to the plane

So the magnetic field at point C will be

$B_C=\left(B_{loop}\right)\hat{i}+\left(B_{wire}\right)\hat{k}$

$B_C=\left(\frac{{\mathrm{μ}}_{0}I}{2R}\right)\hat{i}+\left(\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}R}\right)\hat{k}$

Substitute the values and solve as:

$B_C=\left(\frac{4\mathrm{π}×{10}^{-7}×5.78×{10}^{-3}}{2×0.0189}\right)\hat{i}+\left(\frac{4\mathrm{π}×{10}^{-7}×5.78×{10}^{-3}}{2\mathrm{π}×0.0189}\right)\hat{k}$

$B_C=\left(1.92×{10}^{-7}T\right)\hat{i}+\left(6.12×{10}^{-8}T\right)\hat{k}$