The ratio of magnetic fields at the centre of a current carrying a circular loop of radius 16 cm and at a point on its axis at a distance $r$ from its centre is 125 : 64. The value of $r$ is

12 cm

The correct answer is Option (3) → 12 cm

Magnetic field at the centre of a circular loop:

$B_{centre} = \frac{\mu_0 I}{2R}$

Magnetic field on the axis at distance $r$ from centre:

$B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + r^2)^{3/2}}$

Given ratio:

$\frac{B_{centre}}{B_{axis}} = \frac{125}{64}$

Substitute values:

$\frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I R^2}{2(R^2 + r^2)^{3/2}}} = \frac{125}{64}$

$\frac{(R^2 + r^2)^{3/2}}{R^3} = \frac{125}{64}$

Radius $R = 16 \, \text{cm}$

$(R^2 + r^2)^{3/2} = \frac{125}{64} R^3$

Take power $2/3$:

$R^2 + r^2 = \left(\frac{125}{64}\right)^{2/3} R^2$

$\left(\frac{125}{64}\right)^{2/3} = \left(\frac{5^3}{4^3}\right)^{2/3} = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$

$R^2 + r^2 = \frac{25}{16} R^2$

$r^2 = \frac{25}{16}R^2 - R^2 = \frac{9}{16}R^2$

$r = \frac{3}{4}R$

For $R = 16 \, \text{cm}$:

$r = \frac{3}{4} \times 16 = 12 \, \text{cm}$

Final Answer:

$r = 12 \, \text{cm}$