Consider the situation shown in the figure. The wire $AB$ is sliding on the fixed rails with a constant velocity. If the wire $AB$ is replaced by a semicircular wire,the magnitude of the induced current will

Derive the expression for the mechanical power required to move a conducting rod of length $l$ with a constant velocity $v$ in a uniform magnetic field $B$.

$A$ rectangular loop of length $l$ and breadth $b$ is placed at a distance of $x$ from an infinitely long wire carrying current $i$ such that the direction of the current is parallel to the breadth of the loop. If the loop moves away from the current-carrying wire in a direction perpendicular to it with a velocity $v$,the magnitude of the induced emf in the loop is: ($\mu_0=$ permeability of free space)

The horizontal component of the earth's magnetic field at a place is $3 \times 10^{-4} \ T$ and the dip is $\tan^{-1}(4/3)$. $A$ thin metal rod of length $0.25 \ m$ placed in the north-south position is moved at a constant speed of $10 \ cm/s$ towards the east. Find the $e.m.f.$ induced in the rod across its ends in $\mu V$.

$A$ metal rod of length $l$ rotates about one of its ends in a plane perpendicular to a magnetic field of induction $B$. If the e.m.f. induced between the ends of the rod is $e$,then the number of revolutions made by the rod per second is:

$A$ conducting wire bent in the shape of a semicircle has length $L$ and moves in its plane with constant velocity $v$. $A$ uniform magnetic field $B$ exists in the direction perpendicular to the plane of the wire. The velocity makes an angle $45^{\circ}$ to the diameter joining the free ends,and the emf induced between the ends of the wire is $\Phi = \alpha(B v L)$. The value of the constant $\alpha$ is