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A square coil of area 25 cm^2 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 (1) 2.5 × 10^-3 J (2) 1.0 × 10^-4 J (3) 5 × 10^-3 J (4) 1.0 × 10^-3 J
- 2.5 × 10^-3 J
- 1.0 × 10^-4 J
- 5 × 10^-3 J
- 1.0 × 10^-3 J
Correct answer: 1.0 × 10^-3 J
Solution
The work done in pulling the loop out of the magnetic field is calculated using the formula W = I × V × t, where I is the current induced in the loop, V is the induced voltage, and t is the time. The induced voltage can be found using Faraday's law, and since the resistance is known, the current can be calculated, leading to the correct work done being 1.0 × 10^-3 J.
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