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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2010 · Physics · Electromagnetic Induction

A rectangular loop has a sliding connector PQ of length and...

A rectangular loop has a sliding connector PQ of length ℓ\ell and resistance RΩ\mathrm{R} \Omega and it is moving with a speed vv as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1,I2I_1, I_2 and II are A rectangular loop has a sliding connector PQ of length and resistance R and it is moving with a speed v as shown. The

Options

  1. AI1=−\mathrm{I}_1=-\mathrm{I}2=BℓvR,_2=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}},\mathrm{I}=2 BℓvR=\frac{2 \mathrm{~B} \ell \mathrm{v}}{\mathrm{R}}
  2. BI1=\mathrm{I}_1=\mathrm{I}2=Bℓv3R,_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}},\mathrm{I}=2 Bℓv3R=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}
  3. Cl1=I2=I=BℓvRl_1=I_2=I=\frac{B \ell v}{R}
  4. DI1=I2=Bℓv6R,I=Bℓv3RI_1=I_2=\frac{B \ell v}{6 R}, I=\frac{B \ell v}{3 R}

Correct answer

B. I1=\mathrm{I}_1=\mathrm{I}2=Bℓv3R,_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}},\mathrm{I}=2 Bℓv3R=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}

Step-by-step solution

A moving conductor is equivalent to a battery of emf =vBℓ=\mathrm{v} \mathrm{B} \ell \quad (motion emf) Equivalent circuit \mathrm{I}=\mathrm{l}}_{1}+ \mathrm{l}}_{2} applying Kirchoff's law \begin{aligned}& \mathrm{I}}_{1} R+I R-v B \ell=0 \\ & \mathrm{I}}_{2} R+I R-v B \ell=0 \end{aligned}adding (1) \& (2) \begin{aligned}& 2 \mathrm{IR}+\mathrm{IR}=2 \mathrm{vB} \ell\\ & \mathrm{I}=\frac{2 \mathrm{vB} \ell}{3 \mathrm{R}}\\ & \mathrm{I}}_{1}= \mathrm{I}}_{2}=\frac{\mathrm{vB} \ell}{3 \mathrm{R}} \end{aligned}A rectangular loop has a sliding connector PQ of length and resistance R and it is moving with a speed v as shown. The
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