Q54 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
A conducting circular loop of area $1.0 \mathrm{~m}^{2}$ is placed perpendicular to a magnetic field which varies as $B=\sin (100 t)$ Tesla. If the resistance of the loop is $100 \Omega$, then the average thermal energy dissipated in the loop in one period is $\_\_\_\_$ J.
- A$\pi^{2}$
- Correct answerB$\pi$
- C$\frac{\pi}{2}$
- D$2 \pi$
Solution
Flux: $\phi = BA = \sin(100t)$ Wb
Induced EMF: $\varepsilon = -\frac{d\phi}{dt} = -100\cos(100t)$ V
Induced current: $I = \frac{\varepsilon}{R} = -\cos(100t)$ A
Power dissipated: $P = I^2R = 100\cos^2(100t)$ W
Average power: $P_{avg} = \frac{1}{2} \times 100 = 50$ W
Time period: $T = \frac{2\pi}{100}$ s
Energy in one period: $E = P_{avg} \times T = 50 \times \frac{2\pi}{100} = \pi$ J
Induced EMF: $\varepsilon = -\frac{d\phi}{dt} = -100\cos(100t)$ V
Induced current: $I = \frac{\varepsilon}{R} = -\cos(100t)$ A
Power dissipated: $P = I^2R = 100\cos^2(100t)$ W
Average power: $P_{avg} = \frac{1}{2} \times 100 = 50$ W
Time period: $T = \frac{2\pi}{100}$ s
Energy in one period: $E = P_{avg} \times T = 50 \times \frac{2\pi}{100} = \pi$ J
Hence, the correct answer is option: 'B'.