When a metal rod of length $l$ is placed in a magnetic field $B$ and moved with velocity $v$ perpendicular to the field,then the induced emf across its ends is

An $emf$ of $0.08\,V$ is induced in a metal rod of length $10\,cm$ held normal to a uniform magnetic field of $0.4\,T$,when it moves with a velocity of .......... $m/s$.

$A$ rectangular loop of wire is placed in the $XY$-plane with its side of length $3 \,cm$ parallel to the $X$-axis and the side of length $4 \,cm$ parallel to the $Y$-axis. It is moving in the positive $X$-direction with the speed $10 \,cm/s$. $A$ magnetic field exists in the space with its direction parallel to the $Z$-axis. The field decreases by $2 \times 10^{-3} \,T/cm$ along the positive $X$-axis and increases in time by $2 \times 10^{-2} \,T/s$. The induced emf in the wire is

$A$ metallic square loop $ABCD$ is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced:

As shown in the figure,a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B = 0.15 \, T$. If the resistance is $3 \, \Omega$,the force needed to move the rod as indicated with a constant speed of $2 \, m/s$ is:

$A$ wire of length $L$ is bent into a semicircle and moved with a velocity $v$ in a magnetic field $B$ as shown. What is the induced $emf$ between its two ends?