Two circular coils $1$ and $2$ are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What potential difference in volts should be applied across them so that the magnetic field at the center is the same?
(A) $4$ times of first coil
(B) $6$ times of first coil
(C) $2$ times of first coil
(D) $3$ times of first coil

Hint: It is given that the radius of 1st coil is twice that of the 2nd coil. We know the derivation for finding the magnetic field across a circular coil carrying current. We also know ohm’s law, $V = IR$. From this relation we can submit potential differences in place of current.

Complete step by step solution:
The value of magnetic induction across the axis of the circular coil carrying current is given by
$ \Rightarrow B = \dfrac{{{\mu _0}nI{a^2}}}{{2{{[{a^2} + {x^2}]}^{\dfrac{3}{2}}}}}$
B is the magnetic induction.
${\mu _0}$ is the permeability of free space. \[{\mu _0} = 4\pi \times {10^{ - 7}}H\]
n is the number of turns in the coil.
I is the current flowing through the coil.
a is the radius of the coil.
x is the distance of point P from the center of the coil. The point P is the point where the magnetic field is acting.
Given,
The radius of 1st coil is twice that of 2nd coil
Let the radius of the 1st coil be, ${a_1} = 2r$
Then, the radius of 2nd coil is ${a_2} = r$
The magnetic field at the center of these two coil is equal
We have to find the potential difference which should be applied across them so that their magnetic field at the center of these two coils is equal, V=?
From Ohm’s law we know that
$ \Rightarrow V = IR$
$ \Rightarrow I = \dfrac{V}{R}$
The current in the 1st coil is ${I_1} = \dfrac{{{V_1}}}{{2R}}$ [since resistance is directly proportional to the length]
The current in the 1st coil is ${I_2} = \dfrac{{{V_2}}}{R}$
V is the potential difference
I is the current
R is the resistance
The magnetic field at the center of the 1st coil is given by
 \[ \Rightarrow {B_1} = \dfrac{{{\mu _0}n{I_1}{a_1}^2}}{{2{{[{a_1}^2 + {x^2}]}^{\dfrac{3}{2}}}}}\]
When the magnetic field acts at center then $x = 0$
\[ \Rightarrow {B_1} = \dfrac{{{\mu _0}n\dfrac{{{V_1}}}{{2R}}2{r^2}}}{{2{{[4{r^2} + 0]}^{\dfrac{3}{2}}}}}\]
\[ \Rightarrow {B_1} = \dfrac{{{\mu _0}n\dfrac{{{V_1}}}{{2R}}2{r^2}}}{{8{r^3}}}{\text{ }} \to {\text{1}}\]
The magnetic field at the center of the 2nd coil is given by
\[ \Rightarrow {B_2} = \dfrac{{{\mu _0}n{I_2}{a_2}^2}}{{2{{[{a_2}^2 + {x^2}]}^{\dfrac{3}{2}}}}}\]
When the magnetic field acts at center then $x = 0$
\[ \Rightarrow {B_2} = \dfrac{{{\mu _0}n\dfrac{{{V_2}}}{R}{r^2}}}{{2{{[{r^2} + 0]}^{\dfrac{3}{2}}}}}\]
\[ \Rightarrow {B_2} = \dfrac{{{\mu _0}n\dfrac{{{V_2}}}{R}{r^2}}}{{2{r^3}}}{\text{ }} \to {\text{2}}\]
The magnetic field at the center of these two coils is equal. So, from equation 1 and 2
\[ \Rightarrow \dfrac{{{\mu _0}n\dfrac{{{V_1}}}{{2R}}2{r^2}}}{{8{r^3}}} = \dfrac{{{\mu _0}n\dfrac{{{V_2}}}{R}{r^2}}}{{2{r^3}}}\]
\[ \Rightarrow \dfrac{{{V_1}}}{4} = \dfrac{{{V_2}}}{1}\]
\[ \Rightarrow {V_1} = 4{V_2}\]
The potential difference applied across the 2nd coil is four times of 1st coil.

Hence the correct answer is option (A) $4$ times of first coil

Note: In the solution part we have seen that resistance is directly proportional to the length of the conductor. Resistance of a conductor is directly proportional to the length of the conductor and inversely proportional to its area of cross section.
$ \Rightarrow R \propto \dfrac{L}{A}$
$ \Rightarrow R = \dfrac{{\rho L}}{A}$
R is resistance
$\rho $ is the specific resistance or electrical resistivity
A is the area of the conductor