Find out $V_A - V_B = ?$

$A$ metal conductor of length $1\;m$ rotates vertically about one of its ends at 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 $e.m.f.$ developed between the two ends of the conductor is:

$A$ conducting wire $XY$ of mass $m$ and negligible resistance slides smoothly on two parallel conducting wires as shown in the figure. The closed circuit has a resistance $R$ due to $AC$. $AB$ and $CD$ are perfect conductors. There is a magnetic field $\vec{B} = B(t) \hat{k}$.
$(i)$ Write down the equation for the acceleration of the wire $XY$.
$(ii)$ If $\vec{B}$ is independent of time,obtain $v(t)$,assuming $v(0) = u_0$.
$(iii)$ For $(ii)$,show that the decrease in kinetic energy of $XY$ equals the heat lost in $R$.

$A$ constant force $F$ is applied to a conducting rod of length $l$ moving with constant speed $V$ on two parallel conducting rails connected at the ends by a resistance $R$ in a uniform magnetic field $B$,as shown. If the current flowing through the circuit is $I$,then:

The figure shows an apparatus suggested by Faraday to generate electric current from a flowing river. Two identical conducting plates of length $a$ and width $b$ are placed parallel facing one another on opposite sides of the river flowing with velocity $u$ at a distance $d$ apart. Now both the plates are connected by a load resistance $R$. Then the current through the load $R$ is: (Consider the vertical component of the magnetic field produced by the earth is $B_v$ and the resistivity of river water is $\rho$.)

$A$ metal wire $PQ$ slides on parallel metallic rails having separation $0.25 \ m$,each having negligible resistance. There is a $2 \ \Omega$ resistor and $10 \ V$ battery as shown in the figure. There is a uniform magnetic field directed into the plane of the paper of magnitude $0.5 \ T$. $A$ force of $0.5 \ N$ to the left is required to keep the wire $PQ$ moving with constant speed to the right. With what speed is the wire $PQ$ moving? ..... $m/s$ (Neglect self-inductance of the loop)