Question

Two circular coils $P$ and $Q$ are made from similar wire but radius of $Q$ is twice that of $P$. Relation between the values of potential difference across them so that the magnetic induction at their centers may be the same is :

• A. $V_Q=2V_P$
• B. $V_Q=3V_P$
• C. $V_Q=4V_P$
• D. $V_Q=\frac{1}{4}{V}_P$

Answer 1

Answer: (C)

$V_Q=4V_P$

Magnetic field at the center of a circular coli is $B=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i}{r}$

Where $i$ is flowing in the coil and $r$ is the radius of coil.

At the center of the coil $P$

$B_1=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i_1}{r_1}........\left(1\right)$

At the center of the coil $Q$

$B_2=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i_2}{r_2}........\left(2\right)$

But $B_1=B_2$

$\therefore \frac{{\mu }_0}{4\pi }\times \frac{2\pi i_1}{r_1}=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i_2}{r_2}$

Or $\frac{i_1}{r_1}=\frac{i_2}{r_2}$

As $r_1=2r_2$

$\frac{i_1}{{2r}_2}=\frac{i_2}{r_2}$

Or $i_1=2i_2................\left(3\right)$

Now , Potential differences

$\frac{V_Q}{V_P}=\frac{i_2\times r_2}{i_1\times r_1}=\frac{i_2\times r_2}{{2i}_2\times 2r_2}=\frac{1}{4}\frac{V_P}{V_Q}=\frac{4}{1}{V}_Q=4V_P$

Hence the correct answer is $V_Q=4V_P$