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  • A conducting rod PQ of length L = 1.0 m is moving with a uniform speed in a

A conducting rod PQ of length L = 1.0 m is moving with a uniform speed \mathrm{v=2ms^{-1}} in a uniform magnetic field B = 4.0 T directed into the paper. A capacitor of capacity \mathrm{C=10\mu F} is connected as shown in figure. Then,

 

Option: 1

\mathrm{\mathrm{q}_{\mathrm{A}}=+80 \mu \mathrm{C}\; and\; \mathrm{q}_{\mathrm{B}}=-80 \mu \mathrm{C}}


Option: 2

\mathrm{\mathrm{q}_{\mathrm{A}}=-80 \mu \mathrm{C}\: and \: \mathrm{q}_{\mathrm{B}}=+80 \mu \mathrm{C}}


Option: 3

\mathrm{q}_{\mathrm{A}}=0=\mathrm{q}_{\mathrm{B}}


Option: 4

Charge stored in the capacitor increases exponentially with time


Answers (1)

According to Fleming’s right hand rule, P is at higher potential and Q is at lower potential. Therefore, A is positively charged and B is negatively charged.

\mathrm{\mathrm{Also, charge, \mathrm{Q}=\mathrm{CV}=\mathrm{C}(\mathrm{Bvl}) \begin{aligned} & \quad=10 \times 10^{-6} \times 4 \times 2 \times 1=80 \mu \mathrm{C} \\ & \therefore \quad q_{A-80 \mu C} \text { and } q_B=-80 \mu \mathrm{C} \end{aligned}}}

 

Posted by

Sumit Saini

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