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Electromagnetic Induction question

2010 · Shift 0 · Q57
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Electromagnetic Induction question

2010 · Shift 0 · Q57

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A rectangular loop has a sliding connector PQPQPQ of length lll and resistance RΩR\OmegaRΩ and it is moving with a speed vvv as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1,I2{I_1},{I_2}I1​,I2​ and III are AIEEE 2010 Physics - Electromagnetic Induction Question 115 English
  1. A
    I1=−I2=Blv6R,  I=2Blv6R{I_1} = - {I_2} = {{Blv} \over {6R}},\,\,I = {{2Blv} \over {6R}}I1​=−I2​=6RBlv​,I=6R2Blv​
  2. B
    I1=I2=Blv3R,  I=2Blv3R{I_1} = {I_2} = {{Blv} \over {3R}},\,\,I = {{2Blv} \over {3R}}I1​=I2​=3RBlv​,I=3R2Blv​
  3. C
    I1=I2=I=BlvR{I_1} = {I_2} = I = {{Blv} \over R}I1​=I2​=I=RBlv​
  4. D
    I1=I2=Blv6R,I=Blv3R{I_1} = {I_2} = {{Blv} \over {6R}},I = {{Blv} \over {3R}}I1​=I2​=6RBlv​,I=3RBlv​
View written solutionFree

Correct answer: B

  1. Motional emf across the sliding rod

A rod of length lll moving with speed vvv in a uniform magnetic field BBB has motional emf

E=Blv.\mathcal E = Blv.E=Blv.

This sliding connector PQPQPQ has resistance RRR and behaves like a source of emf BlvBlvBlv in series with resistance RRR.

  1. Equivalent circuit interpretation

The rod connects the top and bottom rails. The rest of the rectangular frame provides two identical paths for current from the top end of the rod to the bottom end:

  • left branch carries current I1I_1I1​
  • right branch carries current I2I_2I2​

Since the frame is symmetric, these two branches are identical. Hence

I1=I2.I_1 = I_2.I1​=I2​.

Let each side branch have resistance RRR (as implied by the standard symmetric setup). Then the two equal resistances RRR and RRR are in parallel, so their equivalent resistance is

Rext=R⋅RR+R=R2.R_{\text{ext}} = \frac{R\cdot R}{R+R} = \frac{R}{2}.Rext​=R+RR⋅R​=2R​.

This external equivalent is in series with the resistance RRR of the moving rod.

Therefore total resistance in the circuit is

Rtotal=R+R2=3R2.R_{\text{total}} = R + \frac{R}{2} = \frac{3R}{2}.Rtotal​=R+2R​=23R​.

  1. Current through the moving rod

The current III through the rod is

= \frac{Blv}{3R/2} = \frac{2Blv}{3R}.$$ --- 4. **Currents in the two side branches** Since the two branches are identical, the rod current splits equally: $$I_1 = I_2 = \frac{I}{2} = \frac{1}{2}\cdot \frac{2Blv}{3R} = \frac{Blv}{3R}.$$ --- 5. **Check with options** We obtained $$I_1 = I_2 = \frac{Blv}{3R}, \qquad I = \frac{2Blv}{3R}.$$ This matches **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.
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