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MHT CET 2019 (2 May, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2019 (2 May, Shift 1)

Q87MHT CET 2019 · 2 MAY · SHIFT 1Easy

Biot-Savart Law & Circular Coils

The magnitude of the 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 given by

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Answer: (b)

For $r \gg R$ : $B = \frac{\mu_0 n I R^2}{2r^3}$ .
With $A = \pi R^2$ : $B = \frac{\mu_0}{4\pi}\cdot\frac{2nIA}{r^3}$ (field of a magnetic dipole $m = nIA$ on its axis).
So the answer is (b).

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