$A$ $2 \; m$ wire is moving with a velocity of $1 \; m/s$ perpendicular to a magnetic field of $0.5 \; Wb/m^2$. The induced e.m.f. in it will be $... \; V$.

$A$ pair of parallel conducting rails lie at a right angle to a uniform magnetic field of $2.0\, T$ as shown in the figure. Two resistors,$10\,\Omega$ and $5\,\Omega$,slide without friction along the rails. The distance between the conducting rails is $0.1\, m$. Then:

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of the Earth's magnetic field in the direction perpendicular to the swing is $B$. The maximum e.m.f. induced across the pendulum is ($g=$ acceleration due to gravity).

$A$ vertical rod of length $l$ is moved with constant velocity $v$ towards the east. If the vertical component of the Earth's magnetic field is $B$ and the angle of dip is $\theta$,then the induced $emf$ in the rod is:

$A$ very small circular loop of radius $a$ is initially (at $t=0$) coplanar and concentric with a much larger fixed circular loop of radius $b$. $A$ constant current $I$ flows in the larger loop. The smaller loop is rotated with a constant angular speed $\omega$ about the common diameter. The emf induced in the smaller loop as a function of time $t$ is

The figure shows a square loop of side $5 \ cm$ being moved towards the right at a constant speed of $1 \ cm/s$. The front edge enters the $20 \ cm$ wide magnetic field $(B = 0.6 \ T)$ at $t = 0$. Find the $emf$ induced in the loop at $(a) \ t = 2 \ s$,$(b) \ t = 10 \ s$,and $(c) \ t = 22 \ s$.