Two insulated circular loop A and B of radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in the figure. The magnitude of the magnetic induction at the centre will be :

Solution

Physics sir · 3 steps

Each loop gives (μ₀ I)/2a at O, but the two loops are perpendicular, so add the fields as vectors.

1. One loop alone. From Biot-Savart, a circular loop of radius a gives a field B = (μ₀ I)/2a at its centre. The field points along the loop's axis, perpendicular to the plane of the loop.
2. Directions at O. Both loops share the centre O, so each gives (μ₀ I)/2a there. Loop A lies horizontal, so its field is vertical. Loop B stands vertical, so its field is horizontal. The two fields are at 90° to each other.

   Bₙₑₜ = √(B_A² + B_B²) = √2 (μ₀ I)/2a = (μ₀ I)/(√2 a)

3. So. The magnetic induction at the centre is (μ₀ I)/(√2 a), which is option (b).

The idea to keep. Field at the centre of a loop points along its axis. With two loops, first find each axis, then add the fields as vectors: perpendicular equal fields give √2 times one of them.

Answer from the NTA final answer key. Solution by PrepWiser.