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  • Solve it, - Electromagnetic Induction and Alternating currents - JEE Main-2

A coil having n turns and resistance R . \Omega is connected with a galvanometer of resistance 4R \Omega. This combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current in the circuit is  

  • Option 1)

    -\frac{W_{2}-W_{1}}{5Rnt}\;

  • Option 2)

    \; \; -\frac{n(W_{2}-W_{1})}{5Rt}\;

  • Option 3)

    \; \; -\frac{(W_{2}-W_{1})}{Rnt}\;

  • Option 4)

    \; \; -\frac{n(W_{2}-W_{1})}{Rt}

 

Answers (1)

best_answer

As we learnt in 

Induced current  I=\frac{-n}{R'}\frac{d\phi }{dt}=\frac{-n}{R'}\frac{dW}{dt}\; where

\phi =W=flux\; \times \; per\; unit\; turn\; of\; the\; coil

\therefore \; \; \; I=-\frac{1}{(R+4R)}\frac{n(W_{2}-W_{1})}{t}=-\frac{n(W_{2}-W_{1})}{5Rt}

Change in flux = W_{2}-W_{1}

 

For N turns -

\varepsilon = \frac{-Nd\phi }{dt}
 

- wherein

N= Number of turns in the Coil 

Negative sign shows that induced emf change the flux.

 

Total current per coil

\therefore I=\frac{\xi }{R_{eq}}=\frac{n}{R_{eq}}\frac{\Delta \phi }{\Delta t }

I=\frac{n(W_{2}-W_{1})}{(R+4R)t}=\frac{n(W_{2}-W_{1})}{5Rt}

Induced current is oppoiste to its cause of production

I=\frac{-n(W_{2}-W_{1})}{5Rt}


Option 1)

-\frac{W_{2}-W_{1}}{5Rnt}\;

This is incorrect option

Option 2)

\; \; -\frac{n(W_{2}-W_{1})}{5Rt}\;

This is correct option

Option 3)

\; \; -\frac{(W_{2}-W_{1})}{Rnt}\;

This incorrect option

Option 4)

\; \; -\frac{n(W_{2}-W_{1})}{Rt}

This incorrect option

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Plabita

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