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  • Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in the figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

 

 

Answers (1)

Due to the motion of the conductor, motional emf is induced in the conductor. It will give rise to a current.

I=\frac{dQ}{dt}=\frac{Bvd}{R}-\frac{Q}{RC}

\frac{Q}{RC}+\frac{dQ}{dt}=\frac{Bvd}{R}

Q+RC\frac{dQ}{dt}=vBCd\; \; \; (Let \; vBdC=A)

Q+RC\frac{dQ}{dt}=A

\frac{dQ}{A-Q}=\frac{1}{RC}dt

By integrating we have

\int_{0}^{Q}\frac{dQ}{A-Q}=\frac{1}{RC}\int_{0}^{t}dt

ln\frac{A-Q}{A}=-\frac{t}{RC}

\frac{A-Q}{A}=e^{-\frac{t}{RC}}

Q=A(1-e^{-\frac{t}{RC}})

Current in the rod

I=\frac{dQ}{dt}=\frac{d}{dt}\left [ A\left ( 1-e^{\frac{-t}{RC}} \right ) \right ]

=-A(e^{\frac{-t}{RC}})\left ( -\frac{1}{RC} \right )

I=\frac{vBd}{R}e^{\frac{-t}{RC}}

 

Posted by

infoexpert23

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