Question: Figure 29-72 shows an arrangement known as a Helmholtz coil. It consists of two circular coaxial coils, each of$\mathbf{200}\mathbf{}\mathbf{turns}$and radius$\mathit{R}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$, separated by a distance$\mathit{s}\mathbf{=}\mathit{R}$. The two coils carry equal currents$\mathit{i}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{mA}$in the same direction. Find the magnitude of the net magnetic field at P, midway between the coils.

The radius of coil is, $R=25\mathrm{cm}=0.25\mathrm{m}$.

The distance between coils is, $s=R=0.25\mathrm{m}$.

The turns in each coils is, $N=200$.

The current in coils is, $i=12.2\mathrm{mA}=0.0122\mathrm{A}$.

The magnetic field of the coil is given by equation 1. But the Helmholtz coil consists of two coils, so the net magnetic field is two times the magnetic field of the coil.

Formula:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{o}}\mathbf{i}{\mathbf{R}}^{\mathbf{2}}\mathbf{n}}{\mathbf{2}{\mathbf{(}{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{Z}}^{\mathbf{2}}\mathbf{)}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}$

From equation $\left(1\right)$,

role="math" localid="1662755468318" $B=\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{iR}}^2\mathrm{n}}{2{({\mathrm{R}}^2+{\mathrm{Z}}^2)}^{{\displaystyle \frac{3}{2}}}}$

We have to find the magnetic field at the point $P$.

Since $s=R$, the point $P$is at a distance $\frac{R}{2}$from the center of the coil.

So, $z=\frac{R}{2}$

Thus, by using the formula

$B=\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{iR}}^2\mathrm{n}}{2{({\mathrm{R}}^2+{{\displaystyle \frac{\mathrm{R}}{2}}}^2)}^{{\displaystyle \frac{3}{2}}}}$

For two coils, the net magnetic field is twice this field.

role="math" localid="1662756071580" $$\begin{gathered} B=\frac{2{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{iR}}^{2}N}{2{({\mathrm{R}}^2+{{\displaystyle \frac{\mathrm{R}}{2}}}^2)}^{{\displaystyle \frac{3}{2}}}} \\ B=\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{iR}}^{2}N}{R^3{(1+{\displaystyle \frac{1}{4}})}^{{\displaystyle \frac{3}{2}}}} \\ B=\frac{{\mathrm{\mu }}_{\mathrm{o}}iN}{R{\left({\displaystyle \frac{5}{4}}\right)}^{{\displaystyle \frac{3}{2}}}} \\ B=\frac{8{\mathrm{\mu }}_{\mathrm{o}}iN}{5\sqrt{5}R} \end{gathered}$$

Substitute all the values in the above equation.

role="math" localid="1662756511182" $$\begin{gathered} B=\frac{8(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{Tm}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.})\left(200\right)(0.0122\mathrm{A})}{5\sqrt{5}(0.25\mathrm{m})} \\ B=8.78\times {10}^{-6}\mathrm{T} \end{gathered}$$

The magnitude of the net magnetic field at P is role="math" localid="1662756497683" $8.78\times {10}^{-6}\mathrm{T}$.