JEE Main 2024 · 27 January · morning · Single correct · +4 / −1
A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is
- (a) -2 V
- (b) +2 V
- (c) +1 V
- (d) -1 V
Answer and solution
Answer (c) +1 V
Solution
Physics sir · 3 steps
Find the flux at the start, note it falls to zero, and let Faraday's minus sign fix the sign of the emf.
- Initial flux. The area is A = 2.5 × 2 = 5 m². The 60° is taken as the angle between B⃗ and the area vector, so Φ₁ = BA cos 60° = 4 × 5 × 1/2 = 10 Wb.
- Final flux. Once the loop is out of the field, no field lines pass through it, so Φ₂ = 0. The change is ΔΦ = 0 - 10 = -10 Wb.
ε_(avg) = -ΔΦ/(Δ t) = -(-10 Wb)/(10 s) = +1 V
- So. The average induced emf is +1 V, which is option (c).
The idea to keep. Average emf is minus the change in flux divided by the time. Work out ΔΦ as final minus initial with its sign, then apply the minus sign from Faraday's law.
Answer from the NTA final answer key. Solution by PrepWiser.