A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is :

Solution

Physics sir · 3 steps

Read the arc angles off the gaps: A spans 270° and B spans 300°, then use B ∝ Iθ/R.

1. Field of an arc. An arc of radius R that subtends angle θ (in radians) at its centre gives a field at the centre proportional to the current and the angle. The radial straight parts point straight at O, so they add nothing there.

   B = (μ₀ I θ)/(4π R)

2. Arc angles. The figure marks the gaps, not the arcs. A leaves a 90° gap, so it covers 360° - 90° = 270° = 3π/2. B leaves a 60° gap, so it covers 300° = 5π/3.

   Wire	I	R	θ	Iθ/R
   A	2 A	2 cm	3π/2	(2 × 3π/2)/2 = 3π/2
   B	3 A	4 cm	5π/3	(3 × 5π/3)/4 = 5π/4

3. So. B_A/B_B = (3π/2)/(5π/4) = 6/5, so the ratio is 6 : 5, option (d).

The idea to keep. For any arc, B = (μ₀ Iθ)/(4π R) with θ the angle the current actually covers. Always check whether the figure marks the arc or the missing gap.

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