Question

The dipole moment of a circular loop carrying a current $I$ , is $m$ and the magnetic field at the centre of the loop is ${B_1}$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is ${B_2}$. The ratio $\dfrac{{{B_1}}}{{{B_2}}}$ is:
(A) $\sqrt 2 $
(B) $\dfrac{1}{{\sqrt 2 }}$
(C) $2$
(D) $\sqrt 3 $

Answer 1

Hint: To solve this question, we have to find the radius when the dipole moment is doubled keeping the current constant. After finding the radius we can use the new radius in the formulae of magnetic field around a current carrying loop and then we can find the ratio of the magnetic field at the centre of the loop with initial radius to that of the new radius.

Formula Used:
Magnitude of a Dipole moment of a circular loop is given as,
$m = IA$
Where $m$is the dipole moment, $I$ is the current in that loop and $A$ is the given area.
Magnetic field for a current carrying loop is given as,
$B = \dfrac{{{\mu _0}I}}{{2\pi R}}$
Where ${\mu _0}$ is the permeability in free space, $I$ is the current in that loop, $R$ is the radius of the loop and $B$is the magnetic field of the loop at the centre.

Complete step by step answer:
In the question the magnitude of magnetic dipole moment of the loop is given as $m$.
But dipole moment is given as,
$m = IA$
Here $A = \pi {R^2}$
$ \Rightarrow m = I\pi {R^2}$
$ \Rightarrow R = \sqrt {\dfrac{m}{{I\pi }}} $
Let the new dipole moment be ${m_r}$. It’s given that the new dipole is twice the dipole at $R$ and the current is the same.
$ \Rightarrow {m_r} = {I_r}\pi R_r^2$
$ \Rightarrow 2{m_{}} = I\pi R_r^2$
$ \Rightarrow {R_r} = \sqrt {\dfrac{{2m}}{{I\pi }}} = \sqrt 2 \times R$
So the radius for the new dipole will be $\sqrt 2 R$.
We already know the formulae for magnetic field for a current carrying loop i.e.
$B = \dfrac{{{\mu _0}I}}{{2\pi R}}$
Where ${\mu _0}$ is the permeability in free space, $I$ is the current in that loop, $R$ is the radius of the loop and $B$ is the magnetic field of the loop at the centre.
$ \Rightarrow $ $B$ is inversely proportional to $R$
So $\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{{R_2}}}{{{R_1}}}$
$ \Rightarrow \dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{\sqrt 2 {R_1}}}{{{R_1}}} = \sqrt 2 $

Hence, the correct option is (A).

Note: In a current carrying loop, whenever the current is kept constant and the distance is varied, the magnetic field changes. It increases with decrease in the radius and decreases with increase in radius. This implies that the magnetic field is maximum at the centre and keeps on decreasing as the distance is increased.