91Ó°ÊÓ

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:

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

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

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

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

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

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

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

Substitute all the values in the above equation.

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

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