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JEE Main 2020 (3 Sep, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2020 (3 Sep, Shift 2)

Q71JEE Main 2020 · 3 SEP · SHIFT 2Medium

Magnetic Flux & Faraday's Law

A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dBdt\frac{dB}{dt} = 0.032 Ts–1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23 ×\times 10–8 Ω\Omegam)

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Answer: (d)

  1. Square side =304=7.5= \frac{30}{4} = 7.5 cm, area A=(0.075)2=5.625×10−3A = (0.075)^2 = 5.625 \times 10^{-3} m2^2.
  2. e=AdBdt=5.625×10−3×0.032=1.8×10−4e = A\frac{dB}{dt} = 5.625 \times 10^{-3} \times 0.032 = 1.8 \times 10^{-4} V
  3. R=ρlπr2=1.23×10−8×0.3π(2×10−3)2≈2.94×10−4 ΩR = \frac{\rho l}{\pi r^2} = \frac{1.23 \times 10^{-8} \times 0.3}{\pi (2 \times 10^{-3})^2} \approx 2.94 \times 10^{-4}\ \Omega
  4. I=eR=1.8×10−42.94×10−4≈0.61I = \frac{e}{R} = \frac{1.8 \times 10^{-4}}{2.94 \times 10^{-4}} \approx 0.61 A
  5. So the answer is (d) 0.61 A.
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