In the given figure,in which direction should the rod be moved to induce an $emf$ between its two ends?

$A$ 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 $R$.

$A$ uniform magnetic field of $0.4 \ \text{T}$ acts perpendicular to a circular copper disc $20 \ \text{cm}$ in radius. The disc is rotating with a uniform angular velocity of $10 \pi \ \text{rad s}^{-1}$ about an axis passing through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim (in $\text{V}$)? $(\pi = 3.14)$

$A$ copper disc of radius $0.1 \ m$ rotates about an axis passing through its centre and perpendicular to its plane with $10 \ \text{revolutions per second}$ in a uniform transverse magnetic field of $0.1 \ T$. The emf induced across the radius of the disc is

$A$ rectangular loop $PQRS$ is being pulled with a constant speed into a uniform transverse magnetic field by a force $F$ (as shown). The $e.m.f.$ induced in side $PS$ and the potential difference between points $P$ and $S$ respectively are (Resistance of the loop $= r$)

$A$ wire of length $L$ having resistance $R$ falls from a height $\ell$ in the Earth's horizontal magnetic field $B$. The induced emf through the wire is ($g$ = acceleration due to gravity).