Q 12-04-238JEE MainJEE Main 2018 (8 Apr)Easy
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:
Answer: (D) $\sqrt2$
$m = I\pi r^2$: doubling $m$ at the same current makes $r' = \sqrt2\,r$. The field at the centre is $B = \dfrac{\mu_0I}{2r}$, so
$$\frac{B_1}{B_2} = \frac{r'}{r} = \sqrt2$$
Solution by Sreeraj P, M.Sc Physics