JEE Main 2023 · 1 February · evening · Single correct · +4 / −1
As shown in the figure, a long straight conductor with semicircular arc of radius π/10m is carrying current I=3A. The magnitude of the magnetic field, at the center O of the arc is :
(The permeability of the vacuum =4π×10⁻⁷ NA⁻²)
- (a) 4μT
- (b) 3μT
- (c) 6μT
- (d) 1μT
Answer and solution
Answer (b) 3μT
Solution
Physics sir · 4 steps
The straight parts point straight at O, so they give no field. Only the semicircle counts.
- Straight wires. Both straight segments lie on the line through O. For every small element dl⃗ on them, dl⃗ is parallel to r⃗, so dl⃗ × r⃗ = 0 and they add nothing to the field at O.
- Semicircular arc. Every element of the arc is at distance R from O and perpendicular to the radius. Half a circle gives half the field of a full loop, (μ₀ I)/2R.
B = (μ₀ I)/4R = (4π × 10⁻⁷ × 3)/(4 × π/10)
- Put in numbers. The π cancels, and the expression becomes (3 × 10⁻⁷)/0.1 = 3 × 10⁻⁶ T.
- So. The field at O is 3 μT, which is option (b).
The idea to keep. A straight wire whose line passes through the point gives zero field there. For an arc of angle θ, use B = (μ₀ I θ)/(4π R), which is (μ₀ I)/4R for a semicircle.
Answer from the NTA final answer key. Solution by PrepWiser.