JEE Main 2023 · 29 January · evening · Single correct · +4 / −1
A square loop of area 25 cm² has a resistance of 10 Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be
- (a) 1.0×10⁻³ J
- (b) 5×10⁻³ J
- (c) 2.5×10⁻³ J
- (d) 1.0×10⁻⁴ J
Answer and solution
Answer (a) 1.0×10⁻³ J
Solution
Physics sir · 4 steps
Pulling slowly means no gain in kinetic energy, so all your work turns into heat in the loop.
- Speed of pulling. Area 25 cm² gives side a = 5 cm = 0.05 m. To take the whole loop out in 1.0 s at constant speed, v = a/t = 0.05 m s⁻¹.
- Motional EMF. Only the edge still inside the field cuts field lines, so e = Bav = 40 × 0.05 × 0.05 = 0.1 V. The current is I = e/R = 0.1/10 = 0.01 A.
- Work equals heat. The speed is constant, so your pull exactly balances the magnetic force on the current. Every joule you do appears as Joule heat, W = e²/R t. Using v = a/t, this becomes:
W = (BA)²/(R t) = (40 × 25 × 10⁻⁴)²/(10 × 1.0) = (0.1)²/10 = 1.0 × 10⁻³ J
- So. The work done is 1.0 × 10⁻³ J, option (a). Check with force: F = BIa = 40 × 0.01 × 0.05 = 0.02 N, and F × a = 0.02 × 0.05 = 1.0 × 10⁻³ J.
The idea to keep. Pulling a loop out uniformly means the work equals the heat produced: W = (BA)²/Rt. Pull it out faster and the work grows, because W goes as 1/t.
Answer from the NTA final answer key. Solution by PrepWiser.