Question

A metal frame $PQRS$ having resistance $2\Omega$, where $PQ=QR=RS=15\text{cm}$ and resistance $2\Omega$ is moved with a speed of $5{\text{cm s}}^{-1}$ in a uniform magnetic field $B=2\text{T}$ which is perpendicular to the plane of the frame as shown in figure. The frame is connected to a network of resistance as shown. Find the current induced in the frame.

• A. $2\text{mA}$
• B. $4\text{mA}$
• C. $5\text{mA}$
• D. $10\text{mA}$

Answer 1

The correct option is C $5\text{mA}$

Equivalent resistance of the network between $P$ and $S$ is given by

$\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{2}+\frac{1}{4}$

$\Rightarrow R_{eq}=1\Omega$

$\therefore$ Total resistance in the circuit will be,

$R=1+2=3\Omega$

Magnitude of emf induced in the frame is

$E=E_{PQ}+E_{QR}+E_{RS}$

$E=0+Blv+0=2\times 0.15\times 5\times {10}^{-2}$

$\therefore E=15\times {10}^{-3}\text{V}$

Induced current in the circuit is given by

$i=\frac{E}{R}=\frac{15\times {10}^{-3}}{3}=5\text{mA}$

Hence, option $\left(\text{C}\right)$ is correct.