A coil of 50 turns and 10 cm diameter has a resistance of 10 $\mathrm{\Omega }$. What must be the potential difference across the coil so as to nullify the earth's magnetic field H = 0.314 G at the center of the coil?

H = 0.314 G = 0.314 x 10^-4 T.          $\left[\because 1 \mathrm{gauss}={10}^{-4} \mathrm{Telsa}\right]$

Let the current in coil be i. Then the magnetic field at the centre of the coil is

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{in}}{2\mathrm{r}}=\frac{4\mathrm{\pi }\times {10}^{-7}\times \mathrm{i}\times 50}{2\times 5\times {10}^{-2}} \\ =2\mathrm{\pi i}\times {10}^{-4}\mathrm{T} \end{gathered}$$

The value of i for which B = H is given by

$2\mathrm{\pi i}\times {10}^{-4} = 0.314\times {10}^{-4}$

or,   i = 0.05 A

$\therefore$ Potential difference = i $\times$R = 0.05 $\times$10 = 0.5 V.