$A$ square loop of side $15 \ cm$ is being moved towards the right at a constant speed of $2 \ cm/s$ as shown in the figure. The front edge enters the $50 \ cm$ wide magnetic field region at $t=0$. The value of the induced emf in the loop at $t=10 \ s$ will be:

$A$ conducting wire of length $2500 \ m$ is kept in the east-west direction at a height of $10 \ m$ from the ground. If it falls freely to the ground,the induced current in the wire is (Resistance of wire $= 25 \sqrt{2} \ \Omega$,acceleration due to gravity $g = 10 \ m/s^2$,$B_H = 2 \times 10^{-5} \ T$). (in $A$)

$A$ square loop of area $25\,cm^2$ has a resistance of $10\,\Omega$. The loop is placed in a uniform magnetic field of magnitude $40.0\,T$. 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 $1.0\,s$ will be $..........\times 10^{-3}\,J$.

Prove that the mechanical power required to move a rod in a uniform magnetic field is converted into electrical power.

$A$ conducting circular loop is rotated about its diameter at a constant angular speed of $100 \ rad/s$ in a magnetic field of $0.5 \ T$ perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced $EMF$ is $15.4 \ mV$. The radius of the loop is . . . . . . $mm$. (Take $\pi = 22/7$)

An aircraft of wing span $40 \ m$ flies horizontally in the Earth's magnetic field of $5 \times 10^{-5} \ T$ at a speed of $500 \ m/s$. The electromotive force (e.m.f.) generated between the tips of the wings of the aircraft is: (in $V$)