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 made three times by keeping the current I same, the magnetic field at the centre of the loop becomes $B_2$. The ratio of $B_1$ and $B_2$ is

$\sqrt{3}:1$

The correct answer is Option (3) → $\sqrt{3}:1$

Magnetic dipole moment of circular loop: $m = I A$, where $A$ is area of loop

Magnetic field at the centre of loop: $B = \frac{\mu_0 I}{2R}$

If dipole moment is made three times: $m_2 = 3 m = I A_2 \Rightarrow A_2 = 3 A_1$

Since $A = \pi R^2 \Rightarrow R_2^2 = 3 R_1^2 \Rightarrow R_2 = \sqrt{3} R_1$

Magnetic field at centre:

$B_1 = \frac{\mu_0 I}{2 R_1}$

$B_2 = \frac{\mu_0 I}{2 R_2} = \frac{\mu_0 I}{2 \sqrt{3} R_1}$

Ratio:

$\frac{B_1}{B_2} = \frac{\mu_0 I / 2 R_1}{\mu_0 I / 2 \sqrt{3} R_1} = \sqrt{3}$

Answer: $B_1 : B_2 = \sqrt{3} : 1$