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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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answer is 6.25.

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Detailed Solution

$\begin{array}{l} m g - B I L = m a \end{array}$

$\begin{array}{l} I = \frac{d}{d t} (q) = \frac{d}{d t} (C E) = \frac{d}{d t} (C B v l) \end{array}$

$[E M F i n d u c e d i n t h e r o d = B v L]$

$\begin{array}{l} \Rightarrow c u r r e n t i n t h e r o d = I = C B L a \end{array}$

$m g = C B^{2} L^{2} a + m a$

$∴ a = \frac{m g}{m + C B^{2} L^{2}} = \frac{(0.1) (10)}{0.1 + (0.015) (4) (1)} = 6.25 m s^{- 2}$

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