In an $AC$ generator, if a coil of $N$ turns and area $A$ is rotated at $v$ revolutions per second in a uniform magnetic field $B$, then the motional $EMF$ produced is equal to (At $t=0$ $s$, the coil is perpendicular to the field).

$A$ thin semicircular conducting ring $(PQR)$ of radius $r$ is falling with its plane vertical in a horizontal magnetic field $B,$ as shown in the figure. The potential difference developed across the ring when its speed is $v,$ is

$A$ copper disc of radius $0.1 \ m$ rotates about its centre with $10$ revolutions per second in a uniform magnetic field of $0.1 \ T$. The emf induced across the radius of the disc is

$A$ conducting bar of mass $m$ and length $l$ moves on two frictionless parallel rails in the presence of a constant uniform magnetic field of magnitude $B$ directed into the page as shown in the figure. The bar is given an initial velocity $v_{0}$ towards the right at $t=0$. Then,the:

Two parallel rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. The distance between the rails is $1 \, m$. $A$ train is travelling with a velocity of $72 \, km/h$ along the track. What is the reading of the millivoltmeter (in $mV$)? (The vertical component of the Earth's magnetic induction is $2 \times 10^{-5} \, T$.)

$A$ 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 $R$.