In a region where the Earth's magnetic field is $3 \times 10^{-4} \, T$ with a dip angle $\theta = \tan^{-1}(4/3)$,a metal rod of length $0.25 \, m$ is placed in the North-South direction. If it is moved towards the East with a velocity of $10 \, cm/s$,calculate the induced $emf$ in $\mu V$.

$A$ long metal bar of $30\,cm$ length is aligned along a north-south line and moves eastward at a speed of $10\,ms^{-1}$. $A$ uniform magnetic field of $4.0\,T$ points vertically downwards. If the south end of the bar has a potential of $0\,V$,the induced potential at the north end of the bar is.....$V$.

The loop shown moves with a velocity $v$ in a uniform magnetic field of magnitude $B$,directed into the paper. The potential difference between $P$ and $Q$ is $e$. Then:

Initially,a rectangular coil with its length vertical is moving out with a constant velocity $v$ in a constant magnetic field $B$ towards the right. Now,the same coil is rotated through $90^{\circ}$ in the same plane in the same magnetic field $B$,and the coil is moving with the same velocity $v$. The magnitude of the induced e.m.f. is now

$A$ wheel with $10$ metallic spokes,each $0.5 \; m$ long,is rotated with a speed of $120 \; rev/min$ in a plane normal to the horizontal component of the Earth's magnetic field $H_{E}$ at a place. If $H_{E} = 0.4 \; G$ at the place,what is the induced $emf$ between the axle and the rim of the wheel? $(1 \; G = 10^{-4} \; T)$

$A$ square frame of side $10\, cm$ and a long straight wire carrying current $1\, A$ are in the plane of the paper. Starting from close to the wire,the frame moves towards the right with a constant speed of $10\, ms^{-1}$ (see figure). The $e.m.f.$ induced at the time the left arm of the frame is at $x = 10\, cm$ from the wire is .....$\mu V$.