Question

Two circular coils $ 1 $ and $ 2 $ are made from the same wire but the radius of the first coil is twice that of the second coil. What potential difference ratio 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)

The magnetic field at the centre of a circular coil,
$B_{\text{centre}} = \frac{\mu_0 Ni}{2r}$
$\because i = \frac{V}{R}$,
and $R = \frac{VA}{\rho\cdot 2 \pi r}$
$\therefore B_{\text{centre}} = \frac{\mu_0 NVA}{4\rho \cdot \pi r^2}$
Where, $A =$ area of cross-section of the wire
According to the question,
$B_1 = B_ 2 = \frac{V_1}{r_1^2} = \frac{V_2}{r_2^2}$
$\Rightarrow \frac{V_1}{(2r_2)^2} = \frac{V_2}{r_2^2}$
$\Rightarrow V_1 = 4V_2$
Ratio of potential difference $ \frac{V_1}{V_2} = 4$