MHT CET · Physics · Electromagnetic Induction
A simple pendulum with bob of mass \(m\) and conducting wire of length L swings under gravity through an angle \(\theta\). The component of earth's magnetic field in the direction perpendicular to swing is B. Maximum e.m.f. induced across the pendulum is ( \(g=\) acceleration due to gravity)
- A \(2 \mathrm{BL}(\sqrt{\mathrm{gL}})(\sin \theta / 2)\)
- B \(\mathrm{BL}(\sqrt{\mathrm{gL}})(\sin \theta / 2)\)
- C \(\mathrm{BL}(\sqrt{\mathrm{gL}})^2(\sin \theta / 2)\)
- D \(2 \mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin ^2 \theta / 2\right)\)
Answer & Solution
Correct Answer
(A) \(2 \mathrm{BL}(\sqrt{\mathrm{gL}})(\sin \theta / 2)\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{2}mv_{max}^2 = mgL(1-\cos\theta) \) \( v_{max} = \sqrt{2gL(1-\cos\theta)} = \sqrt{2gL(2\sin^2(\theta/2))} = 2\sqrt{gL}\sin(\theta/2) \)
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