$A$ conducting rod of length $l$ moves with a constant velocity $\upsilon$ perpendicular to a long,straight wire carrying a current $I$,as shown in the figure. Calculate the $emf$ generated between the ends of the rod.

An infinitely long straight wire carrying current $I$,an open rectangular loop,and a conductor $C$ with a sliding connector are located in the same plane,as shown in the figure. The connector has length $l$ and resistance $R$. It slides to the right with a velocity $v$. The resistance of the conductor and the self-inductance of the loop are negligible. The induced current in the loop,as a function of separation $r$ between the connector and the straight wire,is:

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

$A$ circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ with its plane perpendicular to a uniform magnetic field $B$. Then the emf developed between the centre and the rim of the plate is

$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then:

$A$ straight conductor of length $0.4 \ m$ is moving with a speed of $7 \ ms^{-1}$ perpendicular to a magnetic field of intensity $0.9 \ Wb \ m^{-2}$. The induced emf across the conductor will be (in $V$)