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In the figure shown the section EDGF is fixed. A rod having resistance ' R ' is moved with constant velocity in a uniform magnetic field B as shown in the figure. DE & FG are smooth and resistance less. Initially capacitor is uncharged. The charge on the capacitor: 

Option: 1

remains constant


Option: 2

increases with time                  

              


Option: 3

increases linearly with time                                 


Option: 4

oscillates.


Answers (1)

best_answer

As we have learned 

Motional emf is given by  \varepsilon =BlV

Now applying KVl for EDGF 


 BlV=iR+\frac{q}{C}


or BlV=\left (\frac{dq}{dt} \right )R+\frac{q}{C}

Hence the charge on the capacitor increases with time.

Posted by

seema garhwal

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