Question

A square metal wire loop of side $10\text{cm}$ and resistance $1\Omega$ is moved with a constant velocity $v$ in uniform magnetic field $B=2\text{T}$ as shown in Fig. The magnetic field is perpendicular to the plane of the loop and directed into the paper. The loop is connected to a network of resistors, each equal to $3\Omega$. What should be the speed of the loop so as to have a steady current of $1\text{mA}$ in the loop?

• A. $1\text{cm/s}$
• B. $2\text{cm/s}$
• C. $3\text{cm/s}$
• D. $4\text{cm/s}$

Answer 1

The correct option is B $2\text{cm/s}$

The network $PQRS$ is a balanced Wheatstone's bridge. Hence the resistance of $3\Omega$ between $P$ and $R$ is ineffective. The net resistance of the network, therefore. is $3\Omega$. Total resistance $R=3\Omega +1\Omega =4\Omega$. Now, induced emf is $e=Blv=2\times 0.1\times v=0.2v$.
$\therefore$ Induced currect $I=\frac{e}{R}=\frac{0.2v}{4}$
Given $I=1\times {10}^{-3}\text{A}$
Hence $1\times {10}^{-3}=\frac{0.2v}{4}$
which gives $v=2\times {10}^{-2}\text{m/s}=2\text{cm/s}$, which is choice (b).