Practice portal › Moving Charges and Magnetism › Biot-Savart Law: Straight Wires and Arcs

Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns and carrying current I (ampere) in units of (μ₀ I)/π is

Asked in JEE Main 3rd Sept 1st Shift 2020 · Straight wire segments

Answer: (3) 500√3

Step-by-step solution

Distance from centre to each side d=(√3)/2a=0.05√3 m; each side subtends 30° on either side.

One side: (μ₀ I)/(4π d)(2 sin 30°)=(μ₀ I)/(4π d)

Six sides, 50 turns: B=(300μ₀ I)/(4π(0.05√3))=(μ₀ I)/π·(1500)/(√3)

B=500√3(μ₀ I)/π

Why the other options are wrong

More Biot-Savart Law: Straight Wires and Arcs questionsAll Biot-Savart Law: Straight Wires and Arcs questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer