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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​
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