Two circular coils 1 and 2 are made from the same wire but the radius of the $1^{\text {st }}$ coil is twice…
Two circular coils 1 and 2 are made from the same wire but the radius of the $1^{\text {st }}$ coil is twice that of the $2^{\text {nd }}$ coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same?
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Solution
Let $r_1$ and $r_2$ be the radius of coil 1 and coil 2 . If $B_1$ and $B_2$ are magnetic induction at their centre then $B_1=\frac{\mu_0 I_1}{2 r_1}$ and $B_2=\frac{\mu_0 I_2}{2 r_2}$ $B_1=B_2$ and $r_1=2 r_2$ there $I_1=2 I_2$ Again if $R_1$ and $R_2$ are resistance of the coil 2 then $R_1=2 R_2$ (as $R \propto$ length $=2 \pi r)$ and if $v_1$ and $v_2$ are the potential difference across them respectively then
$\begin{aligned}
\frac{V_1}{V_2} & =\frac{I_1 R_1}{I_2 R_2} \\
& =\frac{\left(2 I_2\right)\left(2 R_2\right)}{I_2 R_2}=4
\end{aligned}$