The loop shown moves with a velocity $v$ in a uniform magnetic field of magnitude $B$,directed into the paper. The potential difference between $P$ and $Q$ is $e$. Then:

Two parallel long straight conductors lie on a smooth surface. Two other parallel conductors rest on them at right angles so as to form a square of side $a$ initially. $A$ uniform magnetic field $B$ exists at right angles to the plane containing the conductors. They all start moving outwards with a constant velocity $v$. If $r$ is the resistance per unit length of the wire,the current in the circuit will be

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

$A$ rigid wire loop of square shape having side of length $L$ and resistance $R$ is moving along the $x$-axis with a constant velocity $v_0$ in the plane of the paper. At $t=0$,the right edge of the loop enters a region of length $3L$ where there is a uniform magnetic field $B$ into the plane of the paper,as shown in the figure. For sufficiently large $v_0$,the loop eventually crosses the region. Let $x$ be the location of the right edge of the loop. Let $v(x)$,$I(x)$,and $F(x)$ represent the velocity of the loop,current in the loop,and force on the loop,respectively,as a function of $x$. Counter-clockwise current is taken as positive. Which of the following schematic plot$(s)$ is(are) correct? (Ignore gravity)

$A$ conducting wire is moving towards the right in a magnetic field $B$. The direction of the induced current $i$ in the wire is shown in the figure. The direction of the magnetic field will be

$A$ rod of length $2 \, m$ slides with a speed of $5 \, ms^{-1}$ on a rectangular conducting frame as shown in the figure. There exists a uniform magnetic field of $0.04 \, T$ perpendicular to the plane of the figure. If the resistance of the rod is $3 \, \Omega$, the current through the rod is