JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Electromagnetic Induction

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An elliptical loop having resistance R, of semi major axis a and semi minor axis b is placed in a 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) $\frac{π ^{2} a ^{2} b ^{2} B ^{2} ω ^{2}}{R}$
(B)zero
(C) $\frac{π ab B ω}{R}$
(D) $\frac{π ^{2} a ^{2} b ^{2} B ^{2} ω ^{2}}{2 R}$ ✓ Correct

Answer: (D)

Solution

As the loop turns in the field the flux through it varies sinusoidally, so the induced emf has amplitude

ε_{0} = B A ω

with

A = π ab

the area of the ellipse, giving

ε_{0} = π ab B ω

. The average power lost to Joule heating is the mean of

ε^{2} / R

over a cycle, and since the mean of

sin^{2}

\frac{1}{2}

P_{avg} = \frac{ε _{0}^{2}}{2 R} = \frac{π ^{2} a ^{2} b ^{2} B ^{2} ω ^{2}}{2 R}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Electromagnetic Induction

Solution

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