Question

A square metal wire loop $P Q R S$ of side $10\, cm$ and resistance $1 \Omega$ is moved with a constant velocity $v$ in a uniform magnetic field of $B=2\, Wb\, m ^{-2}$, 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 $A B C D$ of resistors each of value $3\, \Omega .$ The resistance of the lead wires $S B$ and $R D$ are negligible. The speed of the loop so as to have a steady current of $1\, mA$ in the loop is

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

Answer 1

Answer: (B)

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