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AP EAMCET · PHYSICS · Electromagnetic Induction

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. A \(q_A=+80 \mu \mathrm{C}, q_B=-80 \mu \mathrm{C}\)
  2. B \(q_A=-80 \mu \mathrm{C}, q_B=+80 \mu \mathrm{C}\)
  3. C \(q_A=+1.25 \mu \mathrm{C}, q_B=1.25 \mu \mathrm{C}\)
  4. D \(q_A=-1.25 \mu \mathrm{C}, q_B=+1.25 \mu \mathrm{C}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(q_A=+80 \mu \mathrm{C}, q_B=-80 \mu \mathrm{C}\)

Step-by-step Solution

Detailed explanation

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 \(\pm 80 \mu \mathrm{C}\).