Question

A square metal loop of side $10 \, cm$ and resistance $1\Omega$ is moved with a constant velocity partly inside a uniform magnetic field of $2 \, W b m^{- 2}$ , directed into the paper, as shown in the figure. The loop is connected to a network of five resistors each of value $3\Omega$ . If a steady current of $1 \, mA$ flows in the loop, then the speed of the loop is -

• A. $0.5 \, \text{cm s}^{- 1}$
• B. $1 \, \text{cm s}^{- 1}$
• C. $2 \, \text{cm s}^{- 1}$
• D. $4 \, \text{cm s}^{- 1}$

Answer 1

Answer: (C)

The network appears as a Balanced Wheatstone bridge. Therefore, no current will flow through the bridge resistance. Thus, the resistance will have no effect on the network. Set of two series resistors each values $3 \,\Omega$ is now parallel to each other. Therefore the equivalent resistance of the balanced network is,
$\therefore R=\frac{1}{\left(\frac{1}{3+3}+\frac{1}{3+3}\right)}=\frac{1}{\left(\frac{1}{6}+\frac{1}{6}\right)}=\frac{1}{\left(\frac{2}{6}\right)}=\frac{6}{2}=3 \Omega$
The balanced network is in series with the $1 \Omega$ internal resistance. Therefore, the equivalant resistance of the network is,
$R^{\prime}=3+1=4 \Omega$
The direction of motion of free electron in rod in magnetic fiel $d$ is towards right and it has a velocity $v$. The induced emf is,
$e=v B l=I R^{\prime}$
Therefore, the velocity is,
$v=\frac{I R^{\prime}}{B l}$
$v=\frac{\left(1 \times 10^{-3}\right)(4)}{(2)\left(10 \times 10^{-2}\right)}=2 \times 10^{-2} m s ^{-1}$
$v=2 \,cm s ^{-1}$