If a copper ring is moved quickly towards the south pole of a powerful stationary bar magnet,then:

$A$ conducting circular coil is placed in a uniform magnetic field with the magnetic field initially directed perpendicular to the plane of the coil. In step $A$,the coil is rotated from its initial position by $60^{\circ}$ about its diameter in time $t$. In step $B$,the coil is further rotated about the same axis in the same sense by another $120^{\circ}$ in time $2t$. The ratio of the emf induced in the coil in step $A$ to that in step $B$ is:

Assertion: An $emf$ $\vec{E}$ is induced in a closed loop where magnetic flux is varied. The induced $\vec{E}$ is not a conservative field.
Reason: The line integral $\oint \vec{E} \cdot d\vec{l}$ around the closed loop is nonzero.

Magnetic flux linked with a coil is $\phi = 5t^2 + 2t + 3$,where $t$ is in seconds and $\phi$ is in webers. At time $t = 1 \ s$,the value of the induced emf is . . . . . . $V$.

The magnetic field in a coil of $100$ turns and $40 \text{ cm}^2$ area is increased from $1 \text{ T}$ to $6 \text{ T}$ in $2 \text{ s}$. The magnetic field is perpendicular to the coil. The $e.m.f.$ generated in it is $...... \text{ V}$.

The formula for induced $e.m.f.$ in a coil due to change in magnetic flux through the coil is (here $A$ = area of the coil,$B$ = magnetic field).