A current 'I' flows in anticlockwise direction in a circular arc of a wire having (3/4) th of circumference of a circle of radius R. The magnetic field ' B ' at the centre of circle is ( μ₀ = permeability of free space) (a) (μ₀ I)/(3R) in upward direction (b) (μ₀ I)/(2R) in downward direction (c) (3μ₀ I)/(8R) in downward direction (d) (3μ₀ I)/(8R) in upward direction

Answer: (d). Arc field = (fraction of circle) × (μ₀ I)/(2R). B = 3/4·(μ₀ I)/(2R) = (3μ₀ I)/(8R) Anticlockwise current (seen from above): by the right-hand rule the field at the centre points towards the viewer, i.e. upward. So the answer is (d) (3μ₀ I)/(8R) upward.

A current 'I' flows in anticlockwise direction in a circular arc of a… | MHT CET 2024 (16 May, Shift 2)

Q35MHT CET 2024 · 16 MAY · SHIFT 2Easy

Biot-Savart Law & Circular Coils

A current 'I' flows in anticlockwise direction in a circular arc of a wire having $\left(\frac{3}{4}\right)$ th of circumference of a circle of radius R. The magnetic field ' $B$ ' at the centre of circle is ( $\mu_0 =$ permeability of free space)

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Answer: (d)

Arc field = (fraction of circle) $\times \frac{\mu_0 I}{2R}$ .
$B = \frac{3}{4}\cdot\frac{\mu_0 I}{2R} = \frac{3\mu_0 I}{8R}$
Anticlockwise current (seen from above): by the right-hand rule the field at the centre points towards the viewer, i.e. upward.
So the answer is (d) $\frac{3\mu_0 I}{8R}$ upward.

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Biot-Savart Law & Circular Coils question from MHT CET 2024 (16 May, Shift 2)

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