Q. A rectangular loop has a sliding connector $PQ$ of length $l$ 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) $I_{1} = - I_{2} = \frac{Blv}{R},\ I = \frac{2Blv}{R}$
B) $I_{1} = I_{2} = \frac{Blv}{3R},I = \frac{2Blv}{3R}$
C) $I_{1} = I_{2} = I = \frac{Blv}{R}$
D) $I_{1} = I_{2} = \frac{Blv}{6R},I = \frac{Blv}{3R}$
Explanation
(b) A moving conductor is equivalent to battery of emf
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = vBl$ (motion emf)
Equivalent circuit
$\ \ \ \ \ \ \ \ \ \ \ I = I_{2} + I_{2}$
Applying Kirchhoff’s law
$I_{1}R + IR - vBl = 0\ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$I_{2}R + IR - vBl = 0$ …(ii)
Adding Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2IR + IR = 2vBl$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{2vBl}{3R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I_{1} = I_{2} = \frac{vBl}{3R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.