As a result of a change in the magnetic flux linked to the closed loop as shown in the figure,an $e.m.f.$ $V$ volt is induced in the loop. The work done (in joule) in taking a charge $Q$ coulomb once along the loop is

There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. Then

$A$ 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 $= 10^{-2}\,m^2$).

$A$ metallic ring is dropped down, keeping its plane perpendicular to a constant and horizontal magnetic field. The ring enters the region of magnetic field at $t = 0$ and completely emerges out at $t = T \, \text{sec}$. The current in the ring varies as

$A$ coil having $500$ square loops each of side $10 \ cm$ is placed normal to a magnetic flux which increases at a rate of $1 \ T s^{-1}$. The induced emf is (in $V$)

Magnetic flux linked with the coil is given by $\phi(t) = 2t^2 + 2t + 1$ and its resistance is $10 \ \Omega$. The current passing through the coil at $t = 2 \ s$ is . . . . . . $A$.