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  • a capacitor of capacitance 0.015f is connected to a conducting rail and a rod of mass 100 gm and length 1m it falls under gravity

A capacitor of capacitance C=.015F is connected to parallel conducting rail and a conducting rod of mass m=100g and length l=1m start the fall under gravity in vertical plane. A  uniform magnetic field of 2T exist in space direction perpendicular to rod as shown in figure. Find acceleration of rod. 

                                                               

A) 6.25

B) 1.69

C) 1.44

D) 2.25

 

 

Answers (1)

best_answer

\begin{array}{l} \mathrm{E}=\mathrm{B} \mathrm{lV} \\ \mathrm{Q}=\mathrm{CE} \\ =\mathrm{BlV} \mathrm{C} \\ \mathrm{i}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{BlC} \frac{\mathrm{dV}}{\mathrm{dt}} \\ \Rightarrow \mathrm{F}_{\mathrm{b}}=\mathrm{Bil} \end{array}

\begin{aligned} &=\mathrm{B} \cdot \mathrm{B} \mathrm{lC}\left(\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}\right)^{1}\\ &=\mathrm{B}^{2} \mathrm{l}^{2} \mathrm{C}\left(\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}\right)\\ &\therefore \mathrm{mg}-\mathrm{B}^{2} \mathrm{l}^{2} \mathrm{C} \cdot \mathrm{a}=\mathrm{ma}\\ &\Rightarrow \mathrm{mg}-\left(\mathrm{m}+\mathrm{B}^{2} \mathrm{l}^{2} \mathrm{C}\right) \mathrm{C} \end{aligned}

\large \begin{array}{l} \Rightarrow \mathrm{a}=\frac{\mathrm{mg}}{\mathrm{m}+\mathrm{B}^{2} \mathrm{l}^{2} \mathrm{C}}=\frac{1}{\frac{100}{1000}+4 \times 1^{2} \times \frac{15}{1000}} \\ =\frac{1000}{160} \mathrm{m} / \mathrm{s}^{2} \\ =6.25 \mathrm{m} \end{array}

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avinash.dongre

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