Electromagnetic Induction · Question 18

JEE Main 2026 (Online) 21st January Morning ShiftMCQMedium📈 Topic in 16% papers+4−1

A conducting circular loop of area 1.0 m2 is placed perpendicular to a magnetic field which varies as B = (100t) Tesla. If the resistance of the loop is 100 , then the average thermal energy dissipated in the loop in one period is J.

Answer

(3)

Solution

Flux: = BA = (100t) Wb (A = 1.0 m2)

Induced emf: = - d dt = -100 (100t) V

Power: P = 2R = 104 2(100t)100 = 100 2(100t) W

Period: T = 2 = 2 100 s

Energy in one period: H = 0T P dt = 100 T2 ( 0T 2 t dt = T2 )

H = 50 2 100 = J

Answer: (3)

JEE Main 2026 (Online) 21st January Morning Shift

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