QJEEM23Ph0cc4: JEE Main 2023 Physics - Class 12 JEEM23Ph0cc4 Solution: The ratio of magnetic field at the centre of a current carrying …

The magnetic field at the center of a loop (B₁) is given by

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

The magnetic field on the axis of the loop at a distance ( r ) from the center (B₂) is given by

$B_2 = \frac{\mu_0 Ir^2}{2(r^2 + d^2)^{3/2}}$

where ( d ) is the distance from the center of the coil along the axis. Since ( d = r ), we get

$B_2 = \frac{\mu_0 I}{4\sqrt{2}r}$

The ratio of $B_1$ to $B_2$ is

$\frac{B_1}{B_2} = \frac{\mu_0 I}{2r} \times \frac{4\sqrt{2}r}{\mu_0 I} = \sqrt{8} : 1$

So, the value of ( x ) is 8.