Question

A circular coil of radius \(R\) carries a current \(I\). The magnetic field at its centre is \(B\). At what distance from the centre on the axis of the coil will the magnetic field be \(B/8\)?

• A. \(R\)
• B. \(2R\)
• C. \(\sqrt{3}R\)
• D. \(4R\)

Hint:

Magnetic field on the axis of a current loop decreases with distance according to \[ B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} \] Always compare with the field at the centre to simplify calculations quickly.

Answer 1

The Correct Option is C

Concept: Magnetic field on the axis of a circular current loop at a distance \(x\) from the centre is \[ B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] Magnetic field at the centre of the coil is \[ B_{centre} = \frac{\mu_0 I}{2R} \]

Step 1: Relate the field on axis to the field at the centre. Given \[ B_{axis} = \frac{B_{centre}}{8} \]

Step 2: Substitute the expressions for magnetic fields. \[ \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8}\left(\frac{\mu_0 I}{2R}\right) \] Canceling \( \mu_0 I \) and simplifying, \[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \]

Step 3: Solve for \(x\). \[ (R^2 + x^2)^{3/2} = 8R^3 \] Taking cube root on both sides, \[ R^2 + x^2 = 4R^2 \] \[ x^2 = 3R^2 \] \[ x = \sqrt{3}R \] \[ \boxed{x = \sqrt{3}R} \]