$A$ rod of $10 \ cm$ length is moving perpendicular to a uniform magnetic field of intensity $5 \times 10^{-4} \ Wb/m^2$. If the acceleration of the rod is $5 \ m/s^2$, then the rate of increase of induced $emf$ is . . . . . . .

$A$ jet plane is travelling towards the west at a speed of $1800\, km/h$. What is the voltage difference developed between the ends of the wing having a span of $25\, m$,if the Earth's magnetic field at the location has a magnitude of $5 \times 10^{-4}\, T$ and the dip angle is $30^{\circ}$?

$A$ fighter plane of length $20\, m$, wing span (distance from tip of one wing to the tip of the other wing) of $15\,m$ and height $5\,m$ is flying towards east over Delhi. Its speed is $240\, ms^{-1}$. The earth's magnetic field over Delhi is $5 \times 10^{-5}\,T$ with the declination angle $\, 0^\circ$ and dip of $\theta$ such that $\sin \theta = 2/3$. If the voltage developed is $V_B$ between the lower and upper side of the plane and $V_W$ between the tips of the wings, then $V_B$ and $V_W$ are close to:

$A$ coil of area $80 \, cm^2$ and $50$ turns is rotating with $2000$ revolutions per minute about an axis perpendicular to a magnetic field of $0.05 \, T$. The maximum value of the e.m.f. developed in it is

$A$ 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).

$A$ horizontal telegraph wire of length $30 \ m$ spread east to west falls freely from a height of $20 \ m$. If the resistance of the wire is $40 \ \Omega$ and the horizontal component of the earth's magnetic field at the place is $2 \times 10^{-5} \ T$,then the induced current when the wire reaches the ground is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)