JEE Main 2020 · 3 September · morning · Single correct · +4 / −1
An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency ω, the average power loss in the loop due to Joule heating is :
- (a) (π abBω)/R
- (b) (π ²a²b²B²ω ²)/R
- (c) (π ²a²b²B²ω ²)/2R
- (d) Zero
Answer and solution
Answer (c) (π ²a²b²B²ω ²)/2R
Solution
Physics sir · 4 steps
Flux through the ellipse goes as cos ωt, so the EMF goes as sin ωt, and power averages sin² to one half.
- Flux at time t. The ellipse starts in the xy-plane with area π ab, and B points along z. As it turns about the x-axis, its normal makes angle ω t with B, so Φ = Bπ ab cos ω t.
- Induced EMF. Faraday's law gives the EMF as the rate of change of this flux. It is a sine wave with peak value ε₀ = π abBω.
ε = -dΦ/dt = π abBω sin ω t
- Instantaneous power. Heat generated per second is P = ε²/R = π²a²b²B²ω²/R sin² ω t. Over one full turn, the average of sin² ω t is 1/2.
- So. ⟨ P⟩ = π²a²b²B²ω²/2R, which is option (c).
The idea to keep. For a coil rotating in a uniform field, the EMF is sinusoidal. Power goes as EMF squared, so its average is half the peak value, never zero.
Answer from the NTA final answer key. Solution by PrepWiser.