$A$ wheel with $10$ spokes,each of length $L \ m$,is rotated with a uniform angular velocity $\omega$ in a plane normal to a magnetic field $B$. What is the emf induced between the axle and the rim of the wheel?

$A$ $1\,m$ long metal rod $XY$ completes the circuit as shown in the figure. The plane of the circuit is perpendicular to the magnetic field of flux density $0.15\,T$. If the resistance of the circuit is $5\,\Omega$,the force needed to move the rod in the direction indicated with a constant speed of $4\,m/s$ will be $................\,10^{-3}\,N$.

$A$ metal rod of length $l$ rotates about one of its ends in a plane perpendicular to a magnetic field of induction $B$. If the e.m.f. induced between the ends of the rod is $e$,then the number of revolutions made by the rod per second is:

$A$ square loop of side $15 \ cm$ is being moved towards the right at a constant speed of $2 \ cm/s$ as shown in the figure. The front edge enters the $50 \ cm$ wide magnetic field region at $t=0$. The value of the induced emf in the loop at $t=10 \ s$ will be:

If current $I$ is flowing in a closed circuit with total resistance $R$,the rate of production of heat energy in the loop as we pull it with a constant speed $V$ is ($L=$ length of conductor,$B=$ magnetic field).

$A$ player with a $3 \text{ m}$ long iron rod runs towards the east with a speed of $30 \text{ km/hr}$. The horizontal component of the Earth's magnetic field is $4 \times 10^{-5} \text{ Wb/m}^2$. If the player is running with the rod in horizontal and vertical positions,then the potential difference induced between the two ends of the rod in the two cases will be: