$A$ conducting circular loop is placed in a uniform magnetic field of $0.04\, T$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at a rate of $2\, mm/s$. The induced $emf$ in the loop when the radius is $2\, cm$ is:

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 $B$.

$A$ circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ with its plane perpendicular to a uniform magnetic field $B$. Then the emf developed between the centre and the rim of the plate is

$A$ square loop of area $25 \, cm^2$ has a resistance of $10 \, \Omega$. The loop is placed in a uniform magnetic field of magnitude $40 \, T$. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1 \, s$ will be:

$A$ conducting rod of mass $m$ and length $l$ is free to move without friction on two parallel long conducting rails,as shown in the figure. There is a resistance $R$ across the rails. In the entire space,there is a uniform magnetic field $B$ normal to the plane of the rod and rails. The rod is given an impulsive velocity $v_0$. What happens to the initial kinetic energy $\frac{1}{2} m v_0^2$?

An aeroplane having a wing span of $30 \ m$ flies due north with a speed of $170 \ m/s$. If the vertical component of the Earth's magnetic field is $B = 3.6 \times 10^{-5} \ T$, the potential difference between the tips of the wings will be: (in $mV$)