Question

A square metal wire loop $PQRS$ of side $10cm$ and resistance $1\Omega$ is moved with a constant velocity $v$ in a uniform magnetic field of $B=2Wb{m}^{-2}$, as shown in the 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 $1mA$ in the loop is

• A. $2m{s}^{-1}$
• B. $2\times 10^{-2}m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $200m{s}^{-1}$

Answer 1

Answer: (B)

As the network $ABCD$ is a balanced Wheatstone bridge, no current will flow through $AC$, hence the effective resistance $R$ of bridge is
$\frac{1}{R^{\prime }}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$ or $R^{\prime }=3\Omega$
Total resistance of the circuit, $R=1+3=4\Omega$
Induced emf in the loop, $|\epsilon |=Blv$
Current in the circuit, $I=\frac{|\epsilon |}{R}=\frac{Bvl}{R};v=\frac{IR}{Bl}$
Substituting the given values, we get
$v=\frac{\left(1\times 10^{-3}A\right)\times (4\Omega )}{\left(2Wb{m}^{-2}\right)\times \left(10\times 10^{-2}m\right)}$
$=2\times 10^{-2}m{s}^{-1}$