A circular coil carrying current ' I ' has a radius ' r ' and ' n ' turns. The magnetic field along the axis of a coil at a distance ' 2√(2) r ', from its centre is ( μ₀ = permeability of free space, n is very small) (a) (μ₀ n I)/(9r) (b) (μ₀ n I)/(18r) (c) (μ₀ n I)/(54r) (d) (μ₀ n I)/(27r)

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

A circular coil carrying current ' I ' has a radius ' r ' and ' n '… | MHT CET 2025 (25 Apr, Shift 2)

Q20MHT CET 2025 · 25 APR · SHIFT 2Easy

Biot-Savart Law & Circular Coils

A circular coil carrying current ' I ' has a radius ' $r$ ' and ' $n$ ' turns. The magnetic field along the axis of a coil at a distance ' $2\sqrt{2}\, r$ ', from its centre is ( $\mu_0 =$ permeability of free space, $n$ is very small)

Show answer and solution

Answer: (c)

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

Report a mistake

Biot-Savart Law & Circular Coils question from MHT CET 2025 (25 Apr, Shift 2)

More questions on Biot-Savart Law & Circular Coils