Question

AB and CD are two parallel conductors kept $1$m apart and connected by a resistance R of $6\Omega$ as shown in figure. They are placed in a magnetic field B$=3\times {10}^{-2}$T which is perpendicular to the plane of the conductors and directed into the paper. A wire MN is placed over AB and CD and then made to slide with a velocity $2m{s}^{-1}$.(Neglect the resistance of AB, CD, and MN). Calculate the induced current flowing through the resistor R.

Answer 1

Sliding velocity of wire MN $v=2m/s$

Length of wire MN $l=1m$

Thus rate of change of area of loop MNCA $\frac{dA}{dt}=vl=2m^2/s$

Induced emf in conductor AC $E=B\frac{dA}{dt}=(3\times {10}^{-2})\times 2=6\times {10}^{-2}$ volts

Thus induced current flowing through resistor $I=\frac{E}{R}$

$\therefore$ $I=\frac{6\times {10}^{-2}}{6}=0.01$ A