PX157 - D1d - motional emf

what causes the emf in a moving crossbar?
inside the moving crossbar, there are free charges, $q$ , moving at a velocity, $\vec{v}$ , across the magnetic field, $\vec{B}$
the lorentz force on the conducting (free) charges, $q :$

\vec{F} = q \vec{v} \times \vec{B} = q v B \hat{x} \times (- \hat{z}) = + q v B \hat{y}

positive charges accumulate at point $a$ (top), and negative charges at point $b$ (bottom)
charge separation leads to an electric field:

\vec{E} = - E \hat{y}

electric field creates a force (opposite to the lorentz force):

\vec{F} = - q E \hat{y}

in a steady state, the total force is zero:

\begin{aligned} q \vec{E} + q \vec{v} \times \vec{B} & = \vec{0} \\ \vec{E} = - \vec{v} \times \vec{B} & = - v B \hat{y} \end{aligned}

the voltage between points $a$ and $b :$

V = E L = v L B

the induced electric field (emf) is due to moving charges, the magnetic lorentz force is called a motional emf
emf, $ϵ$ , is the work done per unit charge moving around a closed loop:

\begin{aligned} ϵ & = \frac{W}{q} \\ = \frac{1}{q} \oint \vec{F} \cdot d \vec{l} \\ = \oint (\vec{E} + \vec{v} \times \vec{B}) \cdot d \vec{l} \end{aligned}

the part due to the motion, $\vec{v}$ , is the total motional emf around a closed loop:

ϵ = \oint_{l o o p} (\vec{v} \times \vec{B}) \cdot d \vec{l}