$A$ conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. $A$ uniform magnetic field $\vec{B}$ exists parallel to the axis of rotation. The $e.m.f.$ induced between two ends of the rod is

$A$ conducting wire $XY$ of mass $m$ and negligible resistance slides smoothly on two parallel conducting wires as shown in the figure. The closed circuit has a resistance $R$ due to $AC$. $AB$ and $CD$ are perfect conductors. There is a magnetic field $\vec{B} = B(t) \hat{k}$.
$(i)$ Write down the equation for the acceleration of the wire $XY$.
$(ii)$ If $\vec{B}$ is independent of time,obtain $v(t)$,assuming $v(0) = u_0$.
$(iii)$ For $(ii)$,show that the decrease in kinetic energy of $XY$ equals the heat lost in $R$.

$A$ conducting circular loop is rotated about its diameter at a constant angular speed of $100 \ rad/s$ in a magnetic field of $0.5 \ T$ perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced $EMF$ is $15.4 \ mV$. The radius of the loop is . . . . . . $mm$. (Take $\pi = 22/7$)

$A$ metallic cube of side $15\,cm$ is moving along the $y$-axis at a uniform velocity of $2\,m/s$ in a region of uniform magnetic field of magnitude $0.5\,T$ directed along the $z$-axis. In equilibrium,the potential difference between the faces of higher and lower potential developed because of the motion through the field will be $..........mV$.

When a wire loop is rotated in a magnetic field,the direction of induced $e.m.f.$ changes once in each........$revolution$.

As shown in the figure, a rectangular loop of a conducting wire is moving away with a constant velocity $v$ in a perpendicular direction from a very long straight conductor carrying a steady current $I$. When the breadth of the rectangular loop is very small compared to its distance from the straight conductor, how does the emf $E$ induced in the loop vary with time $t$?