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JEE Mains · Physics · STD 12 -6. Electromagnetic induction

A square loop of side \(10 \mathrm{~cm}\) and resistance \(0.7\  \Omega\) is placed vertically in east-west plane. A uniform magnetic field of \(0.20 \mathrm{~T}\) is set up across the plane in north east direction. The magnetic field is decreased to zero in \(1 \mathrm{~s}\) at a steady rate. Then, magnitude of induced emf is \(\sqrt{\mathrm{x}} \times 10^{-3} \mathrm{~V}\). The value of \(x\) is _______.

  1. A \(1\)
  2. B \(11\)
  3. C \(2\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

\( \overrightarrow{\mathrm{A}}=(0.1)^2 \hat{\mathrm{j}} \) \( \overrightarrow{\mathrm{B}}=\frac{0.2}{\sqrt{2}} \hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}} \hat{\mathrm{j}}\) Magnitude of induced emf…
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