Question

The magnitude of magnetic induction at a point on the axis at a large distance ' \( r \) ' from the centre of a circular coil of ' \( n \) ' turns and area ' \( A \) ' carrying current ' I ' is (\( \mu_0 = \) permeability of free space)

• A. \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{nA}{Ir^3} \)
• B. \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2nIA}{r^3} \)
• C. \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2nI}{Ar^3} \)
• D. \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{nA}{r^3} \)

Hint:

Far from loop: - Treat it as magnetic dipole - $B_{\text{axis}} \propto \frac{1}{r^3}$

Answer 1

The Correct Option is B

Concept: At a large distance, a current loop behaves like a magnetic dipole. Magnetic dipole moment: \[ m = nIA \] Magnetic field on axis of dipole: \[ B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \]

Step 1: Substitute dipole moment.
\[ B = \frac{\mu_0}{4\pi} \frac{2(nIA)}{r^3} \]

Step 2: Simplify.
\[ B = \frac{\mu_0}{4\pi} \frac{2nIA}{r^3} \]