\(v=i\cfrac { \partial x^{ ' } }{ \partial \, t } \)
Using the expression for \(B\) due to a moving charge,
\( \cfrac { \partial B }{ \partial \, t }=-i\cfrac { \partial \, E }{ \partial x^{ ' } } \)
\( \cfrac { \partial B }{ \partial \, t } i\cfrac { \partial x^{ ' } }{ \partial \, t } =-i\cfrac { \partial \, E }{ \partial x^{ ' } }i \cfrac { \partial x^{ ' } }{ \partial \, t }\)
\(\cfrac { \partial B }{ \partial \, t }v=\cfrac { \partial \, E }{ \partial t } \)
\( \int { v \partial B } =\int { \partial \, E } \)
\( |v|.B=E\)
\( q|v|.B=qE=F\)
We have Lorentz force.
But there is a problem.
\(v\) must also be perpendicular to the direction of the moving charge that establish the field for consistent resultant direction. In the case of a circular coil that is always true, as the charge with \(v\) move over the flat end of the coil.
Furthermore, \(E\) is non-existence outside the conductor with a current.
The solution.
\(E\) field has been transformed to \(B\) field. \(E\) field no longer exist, in its place a \(B\) field. The effect of this field however is given by,
\( \cfrac { \partial B }{ \partial \, t }=-i\cfrac { \partial \, E }{ \partial x^{ ' } } \)
as if an \(E\) field exist. Why would \(E\) disappears, yet exist?
Consider an infinitely tight coil that establishes a \(B\) field represented by a single field line. Suppose \(E\) rotates on the plane containing \(v\) and generates a force in the direction of \(-iE\), starting at \(E\) in the downward direction, opposite to \(v\).
When the direction of the parallel component of \(v\) and \(E\) are the same, the force direction reverses and all forces are on the left plane of the circle that defines the coil. The vertical component of the individual forces from the top and bottom quarters cancels. So, the resultant force \(F_L\) (Lorentz's force) is,
\(F_L=4F\int _{ 0 }^{ \pi /2 }{ cos^{ 2 }(\theta ) } d\, \theta=\pi q|v|.B=\pi qE\)
This suggest a \(B\) field point can be equivalently represented as,
irrespective of the sign of the moving charge that established the \(B\) field. This is how \(E\) disappeared but yet still exist. Given any point, the resultant of such a \(E\) distribution is zero, so \(E\) disappeared. A positive charge with a velocity over one such point however, experiences a resultant force, that is \(\pi\) times the force from the interaction of a single \(v\) and \(E\) pair, \(E\) being directly opposite \(v\). The direction of this resultant is in the \(-iE\) direction.
Note: When we note direction of \(v\) and \(x^{'}\) explicitly as in,
\(v=i\cfrac { \partial x^{ ' } }{ \partial \, t } \)
the term \(-i\) disappear from the expression,
\( q|v|.B=qE=F\)
Had we not consider their explicit direction and use,
\(v=\cfrac { \partial x^{ ' } }{ \partial \, t } \)
then we obtain,
\( q|v|.B=-iqE=F\)
the term \(-i\) indicates that \(F\) is in the \(-iE\) direction. Of course we also have \(\pi q(v\times B)=\pi qE=\pi F\) in the case where \(E\) is rotated about \(B\) going directly away. An extra \(\pi\) factor appears in the expression for Lorentz force.
