Question

Suppose you wanted to find the field of a circular loop (Ex. 5.6) at a point r that is not directly above the center (Fig. 5.60). You might as well choose your axes so that r lies in the yz plane at (0, y, z). The source point is (R cos¢', R sin¢', 0), and ¢' runs from 0 to 2Jr. Set up the integrals25 from which you could calculate ${\mathbf{B}}_x$ , ${\mathbf{B}}_y$ and ${\mathbf{B}}_z$and evaluate ${\mathbf{B}}_x$ explicitly.

Short answer

The x, y and z component of the magnetic field of the circular loop are $0$ ,$\frac{{\mu }_{0}IRz}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{\sin \varphi d\varphi }{{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{\frac{3}{2}}}$ and$\frac{{\mu }_{0}IR}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{(R-y\sin \varphi )d\varphi }{{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{3/2}}$ .

Step-by-step solution

Determine the position vector and its magnitude for magnetic field

Consider the figure for the given condition:

The position vector for the magnetic fieldis given as:

$\overrightarrow{\mathbf{p}}\mathbf{=}\mathbf{-}\mathbf{Rcos}\mathit{\varphi }\mathbf{+}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{Rsin}\mathbf{\varphi }\mathbf{)}\widehat{\mathbf{y}}\mathbf{+}\mathbf{z}\widehat{\mathbf{z}}$

Here, $\varphi$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:

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

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

$\begin{array}{c}\overrightarrow{l}=\left(R\cos \varphi \right)\widehat{x}+\left(R\sin \varphi \right)\widehat{y}+z\widehat{z}\\ d\overrightarrow{l}=-\left(R\sin \varphi \right)d\varphi \widehat{x}+\left(R\cos \varphi \right)d\varphi \widehat{y}+0\\ d\overrightarrow{l}=-\left(R\sin \varphi \right)d\varphi \widehat{x}+\left(R\cos \varphi \right)d\varphi \widehat{y}\end{array}$

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

$\begin{array}{l}d\overrightarrow{l}\times \overrightarrow{p}=[-\left(R\sin \varphi \right)d\varphi \widehat{x}+\left(R\cos \varphi \right)d\varphi \widehat{y}]\times [-R\cos \varphi \widehat{x}+(y-R\sin \varphi )\widehat{y}+z\widehat{z}]\\ d\overrightarrow{l}\times \overrightarrow{p}=\left(Rz\cos \varphi d\varphi \right)\widehat{x}+\left(Rz\sin \varphi d\varphi \right)\widehat{y}+(-Ry\sin \varphi d\varphi +R^2+d\varphi )\widehat{z}\end{array}$

Determine the components of magnetic field of circular loop

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

$B_x=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}\left(\frac{d\overrightarrow{l}\times \overrightarrow{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\begin{array}{l}{B}_x=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{[\left(Rz\cos \varphi d\varphi \right)\widehat{x}+\left(Rz\sin \varphi d\varphi \right)\widehat{y}+(-Ry\sin \varphi d\varphi +R^2+d\varphi )\widehat{z}]}{{\left[{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{1/2}\right]}^3}\\ B_x=0\end{array}$

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

$B_y=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}\left(\frac{d\overrightarrow{l}\times \overrightarrow{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\begin{array}{l}{B}_y=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{[\left(Rz\cos \varphi d\varphi \right)\widehat{x}+\left(Rz\right)\widehat{y}+(-Ry\sin \varphi d\varphi +R^2+d\varphi )\widehat{z}]}{{\left[{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{1/2}\right]}^3}\\ B_y=\frac{{\mu }_{0}IRz}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{\sin \varphi d\varphi }{{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{3/2}}\end{array}$

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

$B_z=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}\left(\frac{d\overrightarrow{l}\times \overrightarrow{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\begin{array}{l}{B}_z=\frac{{\mu }_{0}I}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{[\left(Rz\cos \varphi d\varphi \right)\widehat{x}+\left(Rz\right)\widehat{y}+(-Ry+R^2+d\varphi )\widehat{z}]}{{\left[{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{1/2}\right]}^3}\\ B_z=\frac{{\mu }_{0}IR}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{(R-y\sin \varphi )d\varphi }{{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{3/2}}\end{array}$

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

$\frac{{\mu }_{0}IRz}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{\sin \varphi d\varphi }{{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{3/2}}$and $\frac{{\mu }_{0}IR}{4\pi }\underset{0}{\overset{2\pi }{\int }}\frac{(R-y\sin \varphi )d\varphi }{{[{\left(R\cos \varphi \right)}^2+{(y-R\sin \varphi )}^2+z^2]}^{3/2}}$.