JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2024
- Hard
- Numerical
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 _______.
Pick an option to check your answer.
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
\(\mathrm{e}=\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}-0}{1}=\sqrt{2} \times 10^{-3} \mathrm{~V}\)
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