Question

A rectangular loop has a sliding connector PQ of length $\ell$ and resistance $R\,\Omega$ 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. $I_{1} = -I_{2} = \frac{B\ell v}{R}, I = \frac{2B\ell v}{R}$
• B. $I_{1} = -I_{2} = \frac{B\ell v}{3R}, I = \frac{2B\ell v}{3R}$
• C. $I_{1} = -I_{2} = I = \frac{B\ell v}{R}$
• D. $I_{1} = -I_{2} = \frac{B\ell v}{6R}, I = \frac{B\ell v}{3R}$

Answer 1

Answer: (B)

$I_{1} = -I_{2} = \frac{B\ell v}{3R}, I = \frac{2B\ell v}{3R}$

Answer 2

A moving conductor is equivalent to a battery of $emf = v \,B\,\ell$ (motion emf)
Equivalent circuit
$I = I_{1} + I_{2}$
applying Kirchoff?s law
$I_{1}R + IR - vB\ell = 0$ ______(1)
$I_{2}R + IR - vB\ell = 0$ ______(2)
adding $\left(1\right)$ & $\left(2\right)$
$2IR + IR = 2vB\ell$
$I = \frac{2vB\ell}{3R}$
$I_{1} = I_{2} = \frac{vB\ell }{3R}$