Question

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 potentia difference in volts should be applied across them so that the magnetic field at their centres is the same-

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

Magnetic field at the centre of a cicular coil is
$B=\frac{\mu_{0}}{4 \pi} \times \frac{2 \pi i}{r}$
where $i$ is current flowing in the coil and $r$ is radius of coil.
At the centre of coil 1 ,
$B_{1}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \pi i_{1}}{r_{1}} \ldots \ldots(i)$
At the centre of coil 2 ,
$B_{2}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \pi i_{2}}{r_{2}} \ldots \ldots(i i)$
But $B_{1}=B_{2}$
$\therefore \frac{\mu_{0}}{4 \pi} \frac{2 \pi i_{1}}{r_{1}}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i_{2}}{r_{2}}$ or $\frac{i_{1}}{r_{1}}=\frac{i_{2}}{r_{2}}$
As $r_{1}=2 r_{2}$
$\therefore \frac{i_{1}}{2 r_{2}}=\frac{i_{2}}{r_{2}}$ or $i_{1}=2 i_{2}$
Now, ratio of potential differnces
$\frac{V_{2}}{V_{1}}=\frac{i_{1} \times r_{2}}{i_{1} \times r_{1}}=\frac{i_{2} \times r_{2}}{2 i_{2} \times 2 r_{2}}=\frac{1}{4}$
$\therefore \frac{V_{1}}{V_{2}}=\frac{4}{1}$