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Asked in JEE Main 3rd Sept 2nd Shift 2020 · Induced current, heat and power
The wire is 30 cm long, so the square has side 7.5 cm and A=5.625×10⁻³ m².
ε=A(dB)/(dt)=5.625×10⁻³×0.032=1.8×10⁻⁴ V.
Wire cross-section: a=π(2×10⁻³)²=1.257×10⁻⁵ m², so R=(ρ L)/a=(1.23×10⁻⁸×0.3)/(1.257×10⁻⁵)=2.94×10⁻⁴ Ω.
I=(1.8×10⁻⁴)/(2.94×10⁻⁴)=0.61 A.
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