As shown in the figure,a bar magnet is moving towards a stationary coil with a constant speed $v$. The direction of the induced current in the coil as observed by the observer on the $R$.$H$.$S$. is . . . . . . .

$A$ coil of $40 \ \Omega$ resistance has $100$ turns and a radius of $6 \ mm$. It is connected to an ammeter of resistance $160 \ \Omega$. The coil is placed perpendicular to a magnetic field. When the coil is taken out of the field,a charge of $32 \ \mu C$ flows through it. The intensity of the magnetic field is: (in $T$)

The north pole of a magnet is falling towards a metallic ring as shown in the figure. If looked at from above the ring,the direction of the induced current will be:

$A$ coil with $10$ turns and a resistance of $20\,\Omega$ is connected in series with a ballistic galvanometer ($B$.$G$.) of resistance $30\,\Omega$. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction $10^{-2}\,T$. If it is now turned through an angle of $60^{\circ}$ about an axis in its plane, find the charge induced in the coil $..............\times 10^{-5} \, C$ (Area of the coil $= 10^{-2}\,m^2$).

The flux associated with a closed loop is $\phi = 3t^2 + 2t + 5 \text{ Wb}$. If the resistance of the loop is $14 \ \Omega$,then the current induced in this coil at $t = 2 \text{ s}$ is . . . . . . . (in $\text{ A}$)

Derive the relation between induced charge and change in magnetic flux.