$A$ copper disc of radius $0.1 \, m$ is rotated about its centre with $10 \, rev/s$ in a uniform magnetic field of $0.1 \, T$ with its plane perpendicular to the field. The emf induced across the radius of the disc is ........... $V$.

$A$ conducting wire of parabolic shape,initially $y=x^2$,is moving with velocity $\vec{V} = V_0 \hat{i}$ in a non-uniform magnetic field $\vec{B} = B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{k}$,as shown in the figure. If $V_0, B_0, L$ and $\beta$ are positive constants and $\Delta \phi$ is the potential difference developed between the ends of the wire,then the correct statement$(s)$ is/are:
$(1)$ $|\Delta \phi|$ remains the same if the parabolic wire is replaced by a straight wire,$y=x$ initially,of length $\sqrt{2} L$.
$(2)$ $|\Delta \phi|$ is proportional to the length of the wire projected on the $y$-axis.
$(3)$ $|\Delta \phi| = \frac{1}{2} B_0 V_0 L$ for $\beta = 0$.
$(4)$ $|\Delta \phi| = \frac{4}{3} B_0 V_0 L$ for $\beta = 2$.

What is the phase difference between the flux linked with a coil rotating in a uniform magnetic field and the induced e.m.f. produced in it?

Out of the following given loops, in which loop is the direction of the induced current from $a \rightarrow c \rightarrow b$?

$A$ metallic conductor of length $1 \; m$ rotates in a vertical plane parallel to the east-west direction about one of its ends with an angular velocity of $5 \; rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4} \; T$,then the emf induced between the two ends of the conductor is .............

$A$ rectangular loop $PQMN$ with a movable arm $PQ$ of length $12 \, cm$ and resistance $2 \, \Omega$ is placed in a uniform magnetic field of $0.1 \, T$ acting perpendicular to the plane of the loop as shown in the figure. The resistance of the arms $MN$, $NP$, and $MQ$ are negligible. The current induced in the loop when arm $PQ$ is moved with a velocity of $20 \, ms^{-1}$ is (in $ \, A$)