91Ó°ÊÓ

Consider the figure for the given condition:

The position vector for the magnetic fieldis given as:

$\stackrel{\mathbf{→}}{\mathbf{p}}\mathbf{=}\mathbf{-}\mathbf{Rcos}\mathit{ϕ}\mathbf{+}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{Rsin}\mathbf{ϕ}\mathbf{)}\widehat{\mathbf{y}}\mathbf{+}\mathbf{z}\widehat{\mathbf{z}}$

Here, $\mathrm{Ï•}$is the angle for the circular loop and $R$ is the radius of the circular loop.

The magnitude of the position vector is given as:

$\begin{array}{c}p=\sqrt{{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2}\\ p={[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{1/2}\end{array}$

The length of the small element of the circular loop is calculated as:

$\begin{array}{c}\stackrel{→}{l}=\left(R\cos \mathrm{ϕ}\right)\widehat{x}+\left(R\sin \mathrm{ϕ}\right)\widehat{y}+z\widehat{z}\\ d\stackrel{→}{l}=−\left(R\sin \mathrm{ϕ}\right)d\mathrm{ϕ}\widehat{x}+\left(R\cos \mathrm{ϕ}\right)d\mathrm{ϕ}\widehat{y}+0\\ d\stackrel{→}{l}=−\left(R\sin \mathrm{ϕ}\right)d\mathrm{ϕ}\widehat{x}+\left(R\cos \mathrm{ϕ}\right)d\mathrm{ϕ}\widehat{y}\end{array}$

The vector product of position vector and length vector of circular loop is given as:

$\begin{array}{l}d\stackrel{→}{l}×\stackrel{→}{p}=[−\left(R\sin \mathrm{ϕ}\right)d\mathrm{ϕ}\widehat{x}+\left(R\cos \mathrm{ϕ}\right)d\mathrm{ϕ}\widehat{y}]×[−R\cos \mathrm{ϕ}\widehat{x}+(y−R\sin \mathrm{ϕ})\widehat{y}+z\widehat{z}]\\ d\stackrel{→}{l}×\stackrel{→}{p}=\left(Rz\cos \mathrm{ϕ}d\mathrm{ϕ}\right)\widehat{x}+\left(Rz\sin \mathrm{ϕ}d\mathrm{ϕ}\right)\widehat{y}+(−Ry\sin \mathrm{ϕ}d\mathrm{ϕ}+R^2+d\mathrm{ϕ})\widehat{z}\end{array}$

The x component of the magnetic field of circular loop is given as:

$B_x=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\left(\frac{d\stackrel{→}{l}×\stackrel{→}{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\begin{array}{l}{B}_x=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{[\left(Rz\cos \mathrm{ϕ}d\mathrm{ϕ}\right)\widehat{x}+\left(Rz\sin \mathrm{ϕ}d\mathrm{ϕ}\right)\widehat{y}+(−Ry\sin \mathrm{ϕ}d\mathrm{ϕ}+R^2+d\mathrm{ϕ})\widehat{z}]}{{\left[{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{1/2}\right]}^3}\\ B_x=0\end{array}$

The y component of the magnetic field of circular loop is given as:

$B_y=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\left(\frac{d\stackrel{→}{l}×\stackrel{→}{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\begin{array}{l}{B}_y=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{[\left(Rz\cos \mathrm{ϕ}d\mathrm{ϕ}\right)\widehat{x}+\left(Rz\right)\widehat{y}+(−Ry\sin \mathrm{ϕ}d\mathrm{ϕ}+R^2+d\mathrm{ϕ})\widehat{z}]}{{\left[{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{1/2}\right]}^3}\\ B_y=\frac{{\mathrm{μ}}_{0}IRz}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{\sin \mathrm{ϕ}d\mathrm{ϕ}}{{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{3/2}}\end{array}$

The z component of the magnetic field of circular loop is given as:

$B_z=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\left(\frac{d\stackrel{→}{l}×\stackrel{→}{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\begin{array}{l}{B}_z=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{[\left(Rz\cos \mathrm{ϕ}d\mathrm{ϕ}\right)\widehat{x}+\left(Rz\right)\widehat{y}+(−Ry+R^2+d\mathrm{ϕ})\widehat{z}]}{{\left[{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{1/2}\right]}^3}\\ B_z=\frac{{\mathrm{μ}}_{0}IR}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{(R−y\sin \mathrm{ϕ})d\mathrm{ϕ}}{{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{3/2}}\end{array}$

Therefore, the x, y and z component of the magnetic field of the circular loop are $0$,

$\frac{{\mathrm{μ}}_{0}IRz}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{\sin \mathrm{ϕ}d\mathrm{ϕ}}{{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{3/2}}$and $\frac{{\mathrm{μ}}_{0}IR}{4\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\frac{(R−y\sin \mathrm{ϕ})d\mathrm{ϕ}}{{[{\left(R\cos \mathrm{ϕ}\right)}^2+{(y−R\sin \mathrm{ϕ})}^2+z^2]}^{3/2}}$.