JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A conducting circular loop of area \( 1.0 m^{2} \) 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 \( \frac{\pi}{2} \)
- B \( 2\pi \)
- C \( \pi \)
- D \( \pi^{2} \)
Answer & Solution
Correct Answer
(C) \( \pi \)
Step-by-step Solution
Detailed explanation
Area of the loop \(=1 m^2\) \(B=\sin (100 t)\) \(\therefore \quad \phi= BA =\sin (100 t )\) \(\therefore \quad \frac{ d \phi}{ dt }=100 \cos (100 t )\) \(\therefore \quad P=\frac{V^2}{R}=\frac{10^4 \cos ^2(100 t)}{100}\) \(\therefore \quad\) Thermal energy dissipated in 1 time…
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