Q. A rectangular loop with a sliding connector of length $l = 1.0$ m is situated in a uniform magnetic field B = 2T. Perpendicular to the plane of loop. Resistance of connector is $r = 2\Omega$. Two resistance of 6$\ \Omega$ and 3$\ \Omega$ are connected as shown in figure. The external force required to keep the connector moving with a constant velocity $v = 2\ ms^{- 1}$ is
A) 2 N
B) 1 N
C) 4 N
D) 6 N
Explanation
(a) Motion emf induced in the connector
$e = Blv = 2(1)(2) = 4$ V
This acts as a cell of emf 4 V and internal resistance 2$\ \Omega.6\Omega\ $ and 3$\ \Omega$ resistors are in parallel.
$\therefore\ \ \frac{1}{R_{P}} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}\ $
$R_{P} = 2\ \Omega$
$\therefore\ \ \ \ \ $Current through the connector $(i)$
=$\ \frac{E}{R_{P} + r} = \frac{4}{2 + 2} = 1\ $A.
Magnetic force on the connector
$= Bil = (1)(1) = \ 2$ N
Therefore, to keep the connector moving with a constant velocity, a force of 2 N has to be applied to the right side.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.