Consider the situation shown in the figure. The wires $P_1Q_1$ and $P_2Q_2$ are made to slide on the rails with the same speed $5\, cm/s$. Find the electric current in the $9\,\Omega$ resistor if $(a)$ both the wires move towards the right and $(b)$ if $P_1Q_1$ moves towards the left but $P_2Q_2$ moves towards the right.

$A$ horizontal wire is free to slide on the vertical rails of a conducting frame as shown in the figure. The wire has a mass $m$ and length $l$,and the resistance of the circuit is $R$. If a uniform magnetic field $B$ is directed perpendicular to the frame,find the terminal speed of the wire as it falls under the force of gravity.

$A$ square loop of side $a$ and resistance $R$ is moved in a region of uniform magnetic field $B$ (the loop remaining completely inside the field) with a velocity $v$ through a distance $x$. The work done is:

An aircraft with a wingspan of $40 \text{ m}$ flies with a speed of $1080 \text{ km h}^{-1}$ in the eastward direction at a constant altitude in the northern hemisphere,where the vertical component of the earth's magnetic field is $1.75 \times 10^{-5} \text{ T}$. The emf developed between the tips of the wings is: (in $\text{ V}$)

$A$ conducting rod of length $L$ rotates with angular speed $\omega$ in a uniform magnetic field of induction $B$ which is perpendicular to its motion. The induced emf developed between the two ends of the rod is

$A$ 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 $\beta = 2$.