Question

A square metal wire loop $PQRS$ of side $10\text{cm}$ and resistance $1\Omega$ is moved with a constant velocity $v$ in a uniform magnetic field of $B=2{\text{Wbm}}^{-2}$, 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 $SB\text{and}RD$ are negligible. The speed of the loop so as to have a steady current of $1\text{mA}$ in the loop is


$[$1 Mark$]$

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

Answer 1

The correct option is B $2\times {10}^{-2}{\text{ms}}^{-1}$

As the network $ABCD$ is a balanced Wheatstone bridge, no current will flow through $AC$
Hence, the effective resistance $R$ of bridge is

$\frac{1}{R}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}R=3\Omega$

Total resistance of the circuit, $R_{\text{eff.}}=1+3=4\Omega$

Induced emf in the loop,
$\left|\epsilon \right|=Bvl$

Current in the circuit,
$I=\frac{\left|\epsilon \right|}{R}=\frac{Bvl}{R};v=\frac{IR}{Bl}$

Substituting the given values, we get

$v=\frac{(1\times {10}^{-3})\times \left(4\right)}{2\times (10\times {10}^{-2})}=2\times {10}^{-2}{\text{ms}}^{-1}$