$A$ horizontal straight wire $10 \; m$ long extending from east to west is falling with a speed of $5.0 \; m \, s^{-1}$,at right angles to the horizontal component of the earth's magnetic field,$0.30 \times 10^{-4} \; Wb \, m^{-2}$.
$(a)$ What is the instantaneous value of the $emf$ induced in the wire?
$(b)$ What is the direction of the $emf$?
$(c)$ Which end of the wire is at the higher electrical potential?

$A$ 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:

$A$ 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 :

$A$ straight conductor $0.1 \ m$ long moves in a uniform magnetic field of $0.1 \ T$. The velocity of the conductor is $15 \ m/s$ and is directed perpendicular to the field. The emf induced between the two ends of the conductor is: (in $V$)

$A$ current carrying circular loop is perpendicular to a magnetic field of induction $10^{-4} \, T$. If the radius of the loop starts shrinking at a uniform rate of $2 \, mm/s$, then the emf induced in the loop at the instant, when its radius is $20 \, cm$ will be (in $\pi \, \mu V$)

$A$ conducting wire of parabolic shape,initially $y=x^2$,is moving with velocity $\vec{V} = V_0 \hat{i}$ in a non-uniform magnetic field $\vec{B} = B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{k}$,as shown in the figure. If $V_0, B_0, L$ and $\beta$ are positive constants and $\Delta \phi$ is the potential difference developed between the ends of the wire,then the correct statement$(s)$ is/are:
$(1)$ $|\Delta \phi|$ remains the same if the parabolic wire is replaced by a straight wire,$y=x$ initially,of length $\sqrt{2} L$.
$(2)$ $|\Delta \phi|$ is proportional to the length of the wire projected on the $y$-axis.
$(3)$ $|\Delta \phi| = \frac{1}{2} B_0 V_0 L$ for $\beta = 0$.
$(4)$ $|\Delta \phi| = \frac{4}{3} B_0 V_0 L$ for $\beta = 2$.