$A$ metal rod of length '$L$' completes the circuit as shown. The area of the circuit is perpendicular to the magnetic field '$B$'. The total resistance of the circuit is '$R$'. The force needed to move the rod in the direction as shown with a constant speed '$V$' is

$A$ $10\,m$ long horizontal wire extends from North-East to South-West. It is falling with a speed of $5.0\,m/s$,at right angles to the horizontal component of the earth's magnetic field,which is $0.3 \times 10^{-4}\,Wb/m^2$. The value of the induced $emf$ in the wire is:

Two straight conducting plates form an angle $\theta$ where their ends are joined. $A$ conducting bar in contact with the plates and forming an isosceles triangle with them starts at the vertex at time $t=0$ and moves with constant velocity $\vec{v}$ to the right as shown in the figure. $A$ magnetic field $\vec{B}$ points out of the page. The magnitude of the emf induced at $t=1 \text{ s}$ will be

$A$ metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_0 e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

$A$ copper rod of mass $m$ slides under gravity on two smooth parallel rails,with separation $l$ and set at an angle of $\theta$ with the horizontal. At the bottom,the rails are joined by a resistance $R$. There is a uniform magnetic field $B$ normal to the plane of the rails,as shown in the figure. The terminal speed of the copper rod is

$A$ 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: