The dipole moment of a circular loop carrying current \( I \) is \( m \), and the magnetic field at the centre of the loop is \( B_{1} \). When dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is \( B_{2} \). The ratio \( B_{1}/B_{2} \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \sqrt{2} \)
The magnetic moment of a loop is \( m = IA = I\pi R^{2} \), and the field at its centre is \( B = \mu_0 I/2R \), inversely proportional to the radius.
With \( I \) fixed, doubling \( m \) doubles \( R^{2} \):
\( R_2^{2} = 2R_1^{2} \Rightarrow R_2 = \sqrt{2}\,R_1 \)
\( \displaystyle \frac{B_1}{B_2} = \frac{R_2}{R_1} = \sqrt{2} \)