Find out $V_A - V_B = ?$

$A$ conducting rod $AB$ of length $l$ is projected on a frictionless frame $PSRQ$ with velocity $v_0$ in a uniform magnetic field $\vec{B}$ directed into the plane. The velocity of the rod after time $t$ is

$A$ small rectangular loop of wire in the plane of the paper is moved with uniform speed across a limited region of uniform magnetic field perpendicular to the plane of the paper as shown below. Which graph would best represent the variation of the electric current $I$ in the wire with time $t$?

$A$ horizontal straight wire $20 \; m$ long extending from east to west is falling with a speed of $5.0 \; m/s$ at right angles to the horizontal component of the earth's magnetic field $0.30 \times 10^{-4} \; Wb/m^2$. The instantaneous value of the emf induced in the wire will be ......... $mV$.

$A$ circular coil of radius $10 \, cm$ and resistance of $2 \, \Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^{\circ}$ in $0.25 \, s$. If the magnitude of the induced emf is $3.8 \times 10^{-3} \, V$, then the number of turns of the coil is (Horizontal component of earth's magnetic field at the place is $3 \times 10^{-5} \, T$) (in $turns$)

$A$ 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 $\beta = 2$.