$A$ circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ with its plane perpendicular to a uniform magnetic field $B$. Then the emf developed between the centre and the rim of the plate is

$A$ metal disc of radius $R$ rotates with an angular velocity $\omega$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction $B$ acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is (magnitude only):

$A$ rectangular wire loop of sides $5 \text{ cm}$ and $3 \text{ cm}$ with a small cut is moving away from an infinitely long straight wire carrying a current of $30 \text{ A}$ with a velocity of $20 \text{ ms}^{-1}$ as shown in the figure. The emf induced across the cut is (in $\mu V$)

$A$ conductor of length $l$ and mass $m$ can slide along a pair of vertical metal guides connected by a resistance $R$,as shown in the figure. Friction,resistance of the conductor,and guide rails are negligible. There exists a horizontal uniform magnetic field of strength $B$ normal to the plane of the page and directed outward. The terminal speed of fall under the influence of gravity is:

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 2\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\,\mu F$ is connected as shown in the figure. Then

$A$ $1 \ m$ long metal rod $AB$ completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of $0.10 \ T$. If the resistance of the total circuit is $2 \ \Omega$, then the force needed to move the rod towards the right with a constant speed $(v)$ of $1.5 \ m/s$ is . . . . . . $N$.