Q.
A square metal wire loop $PQRS$ of side 10 cm and resistance 1 $\Omega$ is moved with a constant velocity $v_{c}$ in a uniform magnetic field of induction $B = 2\ Wbm^{2}$, as shown in figure. The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to network $ABCD$ of resistors each of value 3$\ \Omega.$ The resistance of the lead wires $SB\ and\ \ RD$ are negligible. The speed of the loop so as to have a steady current of mA in the loop is
A) $2\ ms^{- 1}$
B) $2\ \times 10^{- 2}ms^{- 1}$
C) $20\ ms^{- 1}$
D) $200\ ms^{- 1}$
Explanation
(b) Wheatstone bridge is balanced. Current through $AC$ is zero. Effective resistance R of bridge is
$\frac{1}{R} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3},\ R = 3\Omega$
Total resistance $= 1 + 3 = 4\Omega$
Induced emf
$e = iR = Blv\ $
$\therefore\ \ v = \frac{iR}{Bl} = \ \frac{1 \times 10^{- 3} \times 4}{2 \times 0.1}$
$= 2 \times 10^{- 2}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.