A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be :
(Assume the coil to be short circuited.)

Solution

Physics sir · 4 steps

Halving the turns halves both the emf and the wire length, so track P = ε²/R.

1. Induced emf. The area of each loop does not change, so ε = NAdB/dt and the emf scales as ε ∝ N.
2. Resistance of wire. The wire length is L = N × 2π a, where a is the coil radius, and its cross-section is π r². So R = (ρ N 2π a)/(π r²), which gives R ∝ N/r².

   P = ε²/R ∝ N²/(N/r²) = N r²

3. Put in changes. With N → N/2 and r → 2r, the new power is 1/2 × 4 = 2 times the old one.
4. So. The power dissipated is doubled, option (d).

The idea to keep. For a short-circuited coil, write both ε and R in terms of the turns, because the turns set the wire length. Then P = ε²/R tells you how every change scales.

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