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 ,
$B_1=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i_1}{r_1}\dots \dots (i)$
At the centre of coil 2 ,
$B_2=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i_2}{r_2}\dots \dots (ii)$
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 $r_1=2r_2$
$\therefore \frac{i_1}{2r_2}=\frac{i_2}{r_2}$ or $i_1=2i_2$
Now, ratio of potential differnces
$\frac{V_2}{V_1}=\frac{i_1\times r_2}{i_1\times r_1}=\frac{i_2\times r_2}{2i_2\times 2r_2}=\frac{1}{4}$
$\therefore \frac{V_1}{V_2}=\frac{4}{1}$