A circular conducting coil of radius 1 m is being heated by the change of magnetic field B⃗ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2 μΩ. The magnetic field is slowly switched off such that its magnitude changes in time as
B = 4/π × 10⁻³T( 1 - t/100 )
The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.

Solution

Physics sir · 4 steps

B falls linearly, so the emf is constant: find it once, get the power, then multiply by 100 s.

1. Flux through coil. The area is π r² = π m², so Φ = Bπ = 4 × 10⁻³(1 - t/100) Wb. The π in the area cancels the 1/π in B.
2. Constant emf. By Faraday's law, ε = |dΦ/dt| = (4 × 10⁻³)/100 = 4 × 10⁻⁵ V. It does not depend on t, so the heating power stays the same throughout.

   E = ε²/R t = (4 × 10⁻⁵)²/(2 × 10⁻⁶) × 100

3. Put in numbers. The power is (16 × 10⁻¹⁰)/(2 × 10⁻⁶) = 8 × 10⁻⁴ W. The field reaches zero at t = 100 s, so E = 8 × 10⁻⁴ × 100 = 8 × 10⁻² J.
4. So. The energy dissipated is 0.08 J, which is 80 mJ.

The common slip: 0.08. Reporting the energy in joules gives 0.08, not 80. The blank asks for millijoules, so multiply by 1000.

The idea to keep. If the flux changes linearly with time, the emf is constant. Then heat is just ε²/R times the time it takes the field to reach zero.

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