$A$ conducting bar moves on two conducting rails as shown in the figure. $A$ constant magnetic field $B$ exists into the page. The bar starts to move from the vertex at time $t=0$ with a constant velocity $v$. If the induced $\text{EMF}$ is $E \propto t^n$,then the value of $n$ is . . . . . . .

$A$ wire having mass $m$ and length $l$ can freely slide on a pair of parallel smooth horizontal rails placed in a vertical magnetic field $B$. The rails are connected by a capacitor of capacitance $C$. The electric resistance of the rails and the wire is zero. If a constant force $F$ acts on the wire as shown in the figure,then the acceleration of the wire is given by:

$A$ rod of length $80 \ cm$ rotates about its midpoint with a frequency of $10 \ rev/s$. The potential difference (in volts) between the two ends of the rod due to a magnetic field $B = 0.5 \ T$ directed perpendicular to the rod is:

When a $J$-shaped conducting rod is rotating in its own plane with constant angular velocity $\omega$,about one of its ends $P$,in a uniform magnetic field $\vec B$ directed normally into the plane of the paper,then the magnitude of the emf induced across it will be:

The magnetic induction in the region between the pole faces of an electromagnet is $0.7 \ Wb/m^2$. The induced $e.m.f.$ in a straight conductor $10 \ cm$ long,moving perpendicular to the magnetic field with a velocity of $2 \ m/s$ (where the conductor is also perpendicular to the field and its velocity),is.......$V$.

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.