Question

The magnetic field at the centre of a current carrying circular coil of radius R is \( B_C \) and the magnetic field at a point on its axis at a distance R from its centre is \( B_A \). The value of \( \frac{B_C}{B_A} \) is

• A. \( \sqrt{2} \)
• B. \( \frac{1}{2\sqrt{2}} \)
• C. \( 2\sqrt{2} \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

- Magnetic field at the centre of a circular coil (N turns, radius R, current I): \( B_{centre} = \frac{\mu_0 NI}{2R} \). - Magnetic field on the axis of the coil at distance \(x\) from centre: \( B_{axis} = \frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}} \). - For \(x=R\), \( (R^2+R^2)^{3/2} = (2R^2)^{3/2} = 2^{3/2} (R^2)^{3/2} = 2\sqrt{2} R^3 \).

Answer 1

The Correct Option is C

Magnetic field at the centre of a current-carrying circular coil of radius R and N turns carrying current I: \[ B_C = \frac{\mu_0 N I}{2R} \] (Assuming N=1 if not specified for a single coil).
Magnetic field at a point on the axis of the coil at a distance \(x\) from its centre: \[ B_{axis} = \frac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}} \] Given that the point on the axis is at a distance \(x=R\) from the centre.
So \( B_A \) corresponds to \( B_{axis} \) with \(x=R\).
\[ B_A = \frac{\mu_0 N I R^2}{2(R^2+R^2)^{3/2}} = \frac{\mu_0 N I R^2}{2(2R^2)^{3/2}} \] \[ (2R^2)^{3/2} = (2R^2)\sqrt{2R^2} = 2R^2 \cdot \sqrt{2} R = 2\sqrt{2}R^3 \] So, \( B_A = \frac{\mu_0 N I R^2}{2(2\sqrt{2}R^3)} = \frac{\mu_0 N I R^2}{4\sqrt{2}R^3} = \frac{\mu_0 N I}{4\sqrt{2}R} \).
We need the ratio \( \frac{B_C}{B_A} \): \[ \frac{B_C}{B_A} = \frac{\frac{\mu_0 N I}{2R}}{\frac{\mu_0 N I}{4\sqrt{2}R}} \] \[ \frac{B_C}{B_A} = \frac{\mu_0 N I}{2R} \times \frac{4\sqrt{2}R}{\mu_0 N I} \] Cancel common terms \( \mu_0, N, I, R \): \[ \frac{B_C}{B_A} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \] This matches option (3).