JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 25)
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 $$0.20 T$$ 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 $$\sqrt{x} \times 10^{-3} \mathrm{~V}$$. The value of $$x$$ is __________.
Answer
2
Explanation
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$$\begin{aligned} & \overrightarrow{\mathrm{A}}=(0.1)^2 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{B}}=\frac{0.2}{\sqrt{2}} \hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$$
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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