A conducting rod $P Q$ of length $1 \mathrm{~m}$ is moving with a uniform speed $2 \mathrm{~ms}^{-1}$ in a…

A conducting rod $P Q$ of length $1 \mathrm{~m}$ is moving with a uniform speed $2 \mathrm{~ms}^{-1}$ in a uniform magnetic field of $4 \mathrm{~T}$ which is directed into the paper. A capacitor of capacity $10 \mu \mathrm{F}$ is connected as shown in the figure. Then, the charge on the plates of the capacitor are
  1. $q_A=+80 \mu \mathrm{C}, q_B=-80 \mu \mathrm{C}$
  2. $q_A=-80 \mu \mathrm{C}, q_B=+80 \mu \mathrm{C}$
  3. $q_A=+1.25 \mu \mathrm{C}, q_B=1.25 \mu \mathrm{C}$
  4. $q_A=-1.25 \mu \mathrm{C}, q_B=+1.25 \mu \mathrm{C}$

Solution

Charge on each plate should be $ \begin{aligned} & q=C V \\ & V=v B l \\ & q=C(v B l) \\ & =10 \times 10^{-6} \times 2 \times 4 \times 1 \\ & q=80 \mu \mathrm{C} \\ & \end{aligned} $ where, So, charges on plates are $\pm 80 \mu \mathrm{C}$

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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