What is the cause of the force acting on a sliding conductor placed in a magnetic field?

$A$ $800$ turn coil of effective area $0.05\; m^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{-5}\; T$. When the plane of the coil is rotated by $90^{\circ}$ around any of its coplanar axes in $0.1\; s$,the $emf$ induced in the coil will be.....$V$

$A$ circular coil of radius $30 \ cm$,having $1$ turn and a resistance of ${\pi ^2} \ \Omega$,is placed in a magnetic field of $10^{-2} \ T$. The coil rotates about its diameter with a speed of $200 \ rpm$ in a direction perpendicular to the magnetic field. The value of the induced $AC$ current in the coil will be . . . . . . $mA$.

$A$ conducting rod $AB$ moves parallel to the $X-$ axis in a uniform magnetic field,which is directed perpendicular to the plane of motion (outwards). The end $A$ of the rod gets:

$A$ 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

$A$ conductor wire $ABCDE$ with each arm $10 \ cm$ in length is placed in a magnetic field of $\frac{1}{\sqrt{2}} \ T$,perpendicular to its plane. When the conductor is pulled towards the right with a constant velocity of $10 \ cm/s$,the induced emf between points $A$ and $E$ is . . . . . . $mV$.