$A$ conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. $A$ uniform magnetic field $\vec{B}$ exists parallel to the axis of rotation. The $e.m.f.$ induced between two ends of the rod is

$A$ square conducting frame of side $a$ and a long straight wire carrying current $I$ are placed as shown in the figure. If the frame is moved to the right with a constant velocity $V$,then the induced $emf$ in the frame is proportional to:

When a metal rod of length $l$ is placed in a magnetic field $B$ and moved with velocity $v$ perpendicular to the field,then the induced emf across its ends is

$A$ rectangular conducting loop of length $4 \ cm$ and width $2 \ cm$ is in the $xy$-plane,as shown in the figure. It is being moved away from a thin and long conducting wire along the direction $\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y}$ with a constant speed $v$. The wire is carrying a steady current $I = 10 \ A$ in the positive $x$-direction. $A$ current of $10 \ \mu A$ flows through the loop when it is at a distance $d = 4 \ cm$ from the wire. If the resistance of the loop is $0.1 \ \Omega$,then the value of $v$ is. . . . . . $ms^{-1}$.
[Given: The permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ NA^{-2}$]

$A$ 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 $V$)

When a wire loop is rotated in a magnetic field,the direction of induced $e.m.f.$ changes once in each........$revolution$.