$A$ circular disc of radius $0.1 \ m$ is rotating about its axis at a frequency of $10 \ rev/s$ in a uniform magnetic field of $0.1 \ T$ directed perpendicular to the disc. Calculate the induced $emf$ between the center and the rim of the disc.

Consider a metal ball of radius $r$ moving at a constant velocity $v$ in a uniform magnetic field of induction $\vec{B}$. Assuming that the direction of velocity forms an angle $\alpha$ with the direction of $\vec{B}$,the maximum potential difference between points on the ball is

$XPQY$ is a vertical smooth long loop having a total resistance $R$, where $PX$ is parallel to $QY$ and the separation between them is $l$. $A$ constant magnetic field $B$ perpendicular to the plane of the loop exists in the entire space. $A$ rod $CD$ of length $L$ $(L > l)$ and mass $m$ is made to slide down from rest under gravity as shown in the figure. The terminal speed acquired by the rod is . . . . . . $m/s$. ($g$ = acceleration due to gravity)

$A$ closed loop is moving with constant velocity in a uniform magnetic field as shown. The potential difference and current in element $AB$ will be:

$A$ 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 $t$?

$A$ 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 $\beta = 2$.