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 $\mu V$.

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of the Earth's magnetic field in the direction perpendicular to the swing is $B$. The maximum e.m.f. induced across the pendulum is ($g=$ acceleration due to gravity).

$A$ 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:

The magnetic field in a region is given by $\vec B = B_0(1 + \frac{x}{a})\hat k$. $A$ square loop of edge length $d$ is placed with its edges along the $x$ and $y$ axes. The loop is moved with a constant velocity $\vec V = V_0\hat i$. The $emf$ induced in the loop is:

$A$ conducting rod $PQ$ of length $1 \ m$ is moving with a uniform speed $2 \ ms^{-1}$ in a uniform magnetic field of $4 \ T$ which is directed into the paper. $A$ capacitor of capacity $10 \ \mu F$ is connected as shown in the figure. Then,the charge on the plates of the capacitor are

$A$ $1.0 \ m$ metallic rod is rotated with an angular frequency $200 \ rad \ s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. $A$ constant and uniform magnetic field of $0.5 \ T$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is . . . . . . . (in $V$)