$A$ conducting wire is dropped along the east-west direction. Then,

$A$ metal conductor of length $1\;m$ rotates vertically about one of its ends at an angular velocity of $5\;rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4}\;T$,then the $e.m.f.$ developed between the two ends of the conductor is:

$A$ rectangular wire loop of sides $8 \text{ cm}$ and $3 \text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \text{ T}$ directed normal to the plane of the loop. The emf developed across the cut,if the velocity of the loop is $2 \text{ cm s}^{-1}$ in a direction normal to the shorter side of the loop,will be:

$A$ 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 $\text{ mV}$)

$A$ 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 $R$.

$A$ rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \, \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are: