$A$ square-shaped wire loop of mass $m$,resistance $R$,and side $a$ moving with speed $v_{0}$,parallel to the $X$-axis,enters a region of uniform magnetic field $B$,which is perpendicular to the plane of the loop. The speed of the loop changes with distance $x$ $(x < a)$ in the field as:

$A$ boat is moving due east in a region where the earth's magnetic field is $5.0 \times 10^{-5} \text{ T}$ and is directed due north and horizontal. The boat carries a vertical aerial $2 \text{ m}$ long. If the speed of the boat is $1.5 \text{ m/s}$,calculate the magnitude of the induced $emf$ in the aerial wire in $mV$.

$A$ 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 $R$.

$A$ boat is moving due east in a region where the earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$ due north and horizontal. The boat carries a vertical conducting rod $2 \text{ m}$ long. If the speed of the boat is $2.00 \text{ m/s}$, the magnitude of the induced e.m.f. in the rod is: (in $\text{ mV}$)

$A$ 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 $V$)

The radius of a circular loop placed in a perpendicular uniform magnetic field is increasing at a constant rate of $r_0 \ m s^{-1}$. If at any instant the radius of the loop is $r$,then the emf induced in the loop at that instant will be: