$A$ $1.0 \ m$ metallic rod is rotated with an angular frequency $200 \ rad \ s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. $A$ constant and uniform magnetic field of $0.5 \ T$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is . . . . . . . (in $V$)

$A$ 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 $R$.

Explain the motional $emf$ by the Lorentz force acting on the free charge carriers of a conductor.

$A$ square loop of area $25\,cm^2$ has a resistance of $10\,\Omega$. The loop is placed in a uniform magnetic field of magnitude $40.0\,T$. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0\,s$ will be $..........\times 10^{-3}\,J$.

$A$ long metal rod of length $L$ completes the circuit as shown. The area of the circuit is perpendicular to the magnetic field $B$. The total resistance of the circuit is $R$. The force needed to move the rod in the direction as shown with a constant speed $V$ is:

Statement-$I$: When a conducting rod moves in a uniform transverse magnetic field with a uniform speed which is perpendicular to its length,then a potential difference may develop across its ends.
Statement-$II$: In any conductor,free electrons and free positive ions are available.