$A$ square loop of side $15 \ cm$ is being moved towards the right at a constant speed of $2 \ cm/s$ as shown in the figure. The front edge enters the $50 \ cm$ wide magnetic field region at $t=0$. The value of the induced emf in the loop at $t=10 \ s$ will be:

The loop shown moves with a constant velocity $V$ in a uniform magnetic field of magnetic induction $B$ directed into the paper. The potential difference between $P$ and $Q$ is:

$A$ 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 $\beta = 2$.

$A$ wire loop is rotated in a magnetic field. The frequency of change of direction of the induced $e.m.f.$ is

$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then:

$A$ rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \, \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are: