Question

A conducting rod PQ of length L = 1.0 m is moving with a uniform speed v = 2 m/s in a uniform magnetic field B=4.0 T directed into the paper. A capacitor of capacity $C=10\mu F$ is connected as shown in figure. Then

• A. $qA=+80\mu CandqB=–80\mu C$
  
• B. $qA=–80\mu CandqB=+80\mu C$
  
• C. $qA=0=qB$
• D. Charge stored in the capacitor increases exponentially with time

Answer 1

The correct option is A $qA=+80\mu CandqB=–80\mu C$


$Q=CV=C\left(Bvl\right)=10\times {10}^{–6}\times 4\times 2\times 1=80\mu C$
According to Fleming's right hand rule induced current flows from Q to P. Hence P is at higher potential and Q is at lower potential. Therefore A is positively charged and B is negatively charged.