Magnetic flux and induced emf

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Question

A capacitor of capacitance C = 0.015 F is connected to parallel conducting rail and a conducting rod of mass m = 100 g and length 1m start to fall under gravity in vertical plane. A uniform magnetic field of 2T exist in space directed perpendicular to rod as shown in figure. Find acceleration of rod (m/s²). (use g=10m/s²)

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Solution

$\begin{array}{l} m g - B I L = m a \\ I = \frac{d}{d t} (q) = \frac{d}{d t} (C E) = \frac{d}{d t} (C B v l) E M F i n d u c e d i n t h e r o d = B V L \\ \Rightarrow I = C B L a c u r r e n t i n t h e r o d = I \\ m g = B C B L a L + m a \\ ∴ a = \frac{m g}{m + C B^{2} L^{2}} = \frac{(0.1) (10)}{0.1 + (0.015) (4) (1)} = 6.25 m s^{- 2} \end{array}$

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Similar Questions

The diagram shows a circuit having a coil of resistance $R = 2.5 Ω$ and inductance L connected to a conducting rod PQ which can slide on a perfectly conducting circular ring of radius 10 cm with its centre at 'P'. Assume that friction and gravity are absent and a constant uniform magnetic field of 5 T exists as shown in figure.

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