$A$ straight line conductor of length $0.4 \ m$ is moved with a speed of $7.0 \ ms^{-1}$ perpendicular to a magnetic field of intensity $0.8 \ Wb \ m^{-2}$. The induced e.m.f. across the conductor is (in $V$)

$A$ conducting rod is moving in a uniform magnetic field as shown in the diagram. Which end is at a lower potential?

$A$ constant force $F$ is applied to a conducting rod of length $l$ moving with constant speed $V$ on two parallel conducting rails connected at the ends by a resistance $R$ in a uniform magnetic field $B$,as shown. If the current flowing through the circuit is $I$,then:

$A$ conducting wire bent in the shape of a semicircle has length $L$ and moves in its plane with constant velocity $v$. $A$ uniform magnetic field $B$ exists in the direction perpendicular to the plane of the wire. The velocity makes an angle $45^{\circ}$ to the diameter joining the free ends,and the emf induced between the ends of the wire is $\Phi = \alpha(B v L)$. The value of the constant $\alpha$ is

$A$ copper disc of radius $0.1 \ m$ is rotated about its centre with $10$ revolutions per second in a uniform magnetic field of $0.1 \ T$ with its plane perpendicular to the field. The e.m.f. induced across the radius of the disc is:

$A$ conducting rod $PQ$ of length $5\,m$ oriented as shown in the figure is moving with velocity $(2\,m/s)\hat{i}$ without any rotation in a uniform magnetic field $(3\hat{j} + 4\hat{k})\,T$. The $Emf$ induced in the rod is.....$V$.