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Electromagnetic Induction question

2004 · Shift 0 · Q149
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Electromagnetic Induction question

2004 · Shift 0 · Q149

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A coil having nnn turns and resistance RΩR\OmegaRΩ is connected with a galvanometer of resistance 4RΩ.4R\Omega .4RΩ. This combination is moved in time ttt seconds from a magnetic field W1{W_1}W1​ weber to W2{W_2}W2​ weber. The induced current in the circuit is
  1. A
    (W2−W1)Rnt{{\left( {{W_2} - {W_1}} \right)} \over {Rnt}}Rnt(W2​−W1​)​
  2. B
    −n(W2−W1)5  Rt- {{n\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rt}}−5Rtn(W2​−W1​)​
  3. C
    −(W2−W1)5  Rnt- {{\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rnt}}−5Rnt(W2​−W1​)​
  4. D
    −n(W2−W1)Rt- {{n\left( {{W_2} - {W_1}} \right)} \over {Rt}}−Rtn(W2​−W1​)​
View written solutionFree

Correct answer: B

  1. Given data
  • Coil has nnn turns
  • Resistance of coil =R= R=R
  • Galvanometer resistance =4R= 4R=4R
  • Total resistance in circuit: Rtotal=R+4R=5RR_{\text{total}} = R + 4R = 5RRtotal​=R+4R=5R
  • Magnetic flux changes from W1W_1W1​ to W2W_2W2​ in time ttt

Here, W1W_1W1​ and W2W_2W2​ represent magnetic flux linked with one turn.

  1. Induced emf using Faraday's law

For a coil of nnn turns, e=−nΔWΔte = -n\frac{\Delta W}{\Delta t}e=−nΔtΔW​ where ΔW=W2−W1\Delta W = W_2 - W_1ΔW=W2​−W1​ So, e=−nW2−W1te = -n\frac{W_2 - W_1}{t}e=−ntW2​−W1​​

  1. Induced current

Using Ohm’s law, i=eRtotali = \frac{e}{R_{\text{total}}}i=Rtotal​e​ Substitute values: i=−nW2−W1t5Ri = \frac{-n\dfrac{W_2 - W_1}{t}}{5R}i=5R−ntW2​−W1​​​

Therefore, i=−n(W2−W1)5Rti = -\frac{n(W_2 - W_1)}{5Rt}i=−5Rtn(W2​−W1​)​

  1. Match with options

This matches Option B.

  1. Final answer

i=−n(W2−W1)5Rt\boxed{i=-\frac{n(W_2-W_1)}{5Rt}}i=−5Rtn(W2​−W1​)​​

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