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?

$A$ conducting rod $PQ$ of length $L = 1.0 \, m$ is moving with a uniform speed $v = 2 \, m/s$ in a uniform magnetic field $B = 4.0 \, T$ directed into the paper. $A$ capacitor of capacity $C = 10 \, \mu F$ is connected as shown in the figure. Then:

$A$ metal rod $AB$ of length $50 \text{ cm}$ is moving at a velocity $8 \text{ ms}^{-1}$ in a magnetic field of $2 \text{ T}$. If the field is at $60^{\circ}$ with the plane of motion as shown in the figure,then the potentials $V_A$ and $V_B$ are related by

There is a uniform magnetic field $B$ normal to the $xy$ plane. $A$ conductor $ABC$ has length $AB = l_1$,parallel to the $x$-axis,and length $BC = l_2$,parallel to the $y$-axis. $ABC$ moves in the $xy$ plane with velocity $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The potential difference between $A$ and $C$ is proportional to

$A$ $1.0 \ m$ metallic rod is rotated with an angular frequency $200 \ rad \ s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. $A$ constant and uniform magnetic field of $0.5 \ T$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is . . . . . . . (in $V$)

In the given branch $AB$ of a circuit,a current $I = (10t + 5) \, A$ is flowing,where $t$ is time in seconds. At $t = 0$,the potential difference between points $A$ and $B$ $(V_A - V_B)$ is.....$V$.