A rectangular loop has a sliding connector PQ of length $\ell$ and resistance $\mathrm{R} \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are

• A

  $\mathrm{I}_1=-\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{\mathrm{R}}$
• Correct answer
  B

  $\mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}$
• C

  $l_1=I_2=I=\frac{B \ell v}{R}$
• D

  $I_1=I_2=\frac{B \ell v}{6 R}, I=\frac{B \ell v}{3 R}$

Solution

A moving conductor is equivalent to a battery of emf $=\mathrm{v} \mathrm{B} \ell \quad$ (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) $$ \begin{aligned} & 2 \mathrm{IR}+\mathrm{IR}=2 \mathrm{vB} \ell \\ & \mathrm{I}=\frac{2 \mathrm{vB} \ell}{3 \mathrm{R}} \\ & \mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{vB} \ell}{3 \mathrm{R}} \end{aligned} $$

Hence, the correct answer is option: 'B'.