$A$ rectangular loop is provided with a sliding connector of length $1 \, m$ and resistance $2 \, \Omega$. It is placed in a uniform magnetic field of $2 \, T$ perpendicular to the plane of the loop. The external force required to keep the connector moving with a uniform velocity of $2 \, ms^{-1}$ is: (in $N$)

Two identical conductors $P$ and $Q$ are placed on two frictionless rails $R$ and $S$ in a uniform magnetic field directed into the plane. If $P$ is moved in the direction shown in the figure with a constant speed,then rod $Q$

$A$ bicycle wheel of radius $0.4\, m$ has $20$ spokes. It is rotating at the rate of $180$ revolutions per minute, perpendicular to the horizontal component of the Earth's magnetic field of $0.4 \times 10^{-4}\, T$. The $emf$ induced between the rim and the centre of the wheel will be

$A$ wire of length $1 \, m$ is moving at a speed of $2 \, ms^{-1}$ perpendicular to its length and a homogeneous magnetic field of $0.5 \, T$. The ends of the wire are joined to a circuit of resistance $6 \, \Omega$. The rate at which work is being done to keep the wire moving at constant speed is:

$A$ very small circular loop of radius $a$ is initially (at $t=0$) coplanar and concentric with a much larger fixed circular loop of radius $b$. $A$ constant current $I$ flows in the larger loop. The smaller loop is rotated with a constant angular speed $\omega$ about the common diameter. The emf induced in the smaller loop as a function of time $t$ is

$A$ wheel with $10$ metallic spokes,each $0.5 \; m$ long,is rotated with a speed of $120 \; rev/min$ in a plane normal to the horizontal component of the Earth's magnetic field $H_{E}$ at a place. If $H_{E} = 0.4 \; G$ at the place,what is the induced $emf$ between the axle and the rim of the wheel? $(1 \; G = 10^{-4} \; T)$