$A$ metallic rod of length $20 \,cm$ is placed in the North-South direction and is moved at a constant speed of $20 \,m/s$ towards the East. The horizontal component of the Earth's magnetic field at that place is $4 \times 10^{-3} \,T$ and the angle of dip is $45^{\circ}$. The emf induced in the rod is ............. $mV$.

Prove that the mechanical power required to move a rod in a uniform magnetic field is converted into electrical power.

Two straight conducting plates form an angle $\theta$ where their ends are joined. $A$ conducting bar in contact with the plates and forming an isosceles triangle with them starts at the vertex at time $t=0$ and moves with constant velocity $\vec{v}$ to the right as shown in the figure. $A$ magnetic field $\vec{B}$ points out of the page. The magnitude of the emf induced at $t=1 \text{ s}$ will be

$A$ conducting rod of length $l$ moves with velocity $\upsilon$ in a direction parallel to a long wire carrying a steady current $I$. The axis of the rod is maintained perpendicular to the wire with the near end at a distance $r$ away as shown in the figure. Find the emf induced in the rod.

$A$ conducting circular loop is placed in a uniform magnetic field,$B = 0.025 \, T$,with its plane perpendicular to the field. The radius of the loop is made to shrink at a constant rate of $1 \, mm \, s^{-1}$. The induced $emf$ when the radius is $2 \, cm$ is:

$A$ conductor wire $ABCDE$ with each arm $10 \ cm$ in length is placed in a magnetic field of $\frac{1}{\sqrt{2}} \ T$,perpendicular to its plane. When the conductor is pulled towards the right with a constant velocity of $10 \ cm/s$,the induced emf between points $A$ and $E$ is . . . . . . $mV$.