$A$ copper rod $AB$ of length $L$,pivoted at one end $A$,rotates at a constant angular velocity $\omega$,at right angles to a uniform magnetic field of induction $B$. The e.m.f. developed between the midpoint $C$ of the rod and end $B$ is

$A$ 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:

$A$ metal conductor of length $1 \ m$ rotates vertically about one of its ends at an angular velocity of $5 \ rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4} \ T$,then the e.m.f. developed between the two ends of the conductor is:

$A$ rod of mass $m$ and resistance $R$ slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $\theta$ with respect to the horizontal as shown in the figure. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field $\vec{B}$ along the vertical direction. If the rod is initially at rest,find the velocity of the rod as a function of time.

$A$ 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 . . . . . . .

$A$ rectangular wire loop of sides $8 \text{ cm}$ and $3 \text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \text{ T}$ directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is $2 \text{ cm s}^{-1}$ in a direction normal to the shorter side of the loop, will be: