A circular coil having n turns, each of radius 'r' carries a current I. The magnetic field at a distance x = √(8) r from its centre along the axis of the coil is ( μ₀ = permeability of free space) (a) (μ₀ n I)/(51 r) (b) (μ₀ n I)/(52 r) (c) (μ₀ n I)/(53 r) (d) (μ₀ n I)/(54 r)

Answer: (d). Axial field: B = μ₀ n I r²2(r²+x²)3/2 x² = 8r² ⇒ (r²+x²)3/2 = (9r²)3/2 = 27r³ B = (μ₀ n I r²)/(2× 27 r³) = (μ₀ n I)/(54 r) So the answer is (d).

A circular coil having n turns, each of radius 'r' carries a current… | MHT CET 2026 (20 May, Shift 1)

Q1MHT CET 2026 · 20 MAY · SHIFT 1Easy

Biot-Savart Law & Circular Coils

A circular coil having n turns, each of radius 'r' carries a current I. The magnetic field at a distance $x = \sqrt{8}\, r$ from its centre along the axis of the coil is ( $\mu_0$ = permeability of free space)

Show answer and solution

Answer: (d)

Axial field: $B = \frac{\mu_0 n I r^2}{2(r^2+x^2)^{3/2}}$
$x^2 = 8r^2 \Rightarrow (r^2+x^2)^{3/2} = (9r^2)^{3/2} = 27r^3$
$B = \frac{\mu_0 n I r^2}{2\times 27 r^3} = \frac{\mu_0 n I}{54 r}$
So the answer is (d).

Report a mistake

Biot-Savart Law & Circular Coils question from MHT CET 2026 (20 May, Shift 1)

More questions on Biot-Savart Law & Circular Coils