How to derive the curl of E equation in the frame of the conductor?

[adelmakram]

According to wikipedia, "The moving magnet and conductor problem", I stopped at the equation shown in the attachment.
It said that the curl of the E` ( electric field in the frame of the conductor) is equal to minus of the dot product of the velocity of the conductor and the del multiplied by the magnetic field.

How to derive this formula?

 

[UltrafastPED]

The first part is follows from the chain rule: substituting B(x'+vt) for B':

∂B'/∂t = ∂B(x'+vt)/∂t = ∇B(x') ° ∂vt/∂t = ∇B(x') ° v.

But this can also be expressed as (v°∇)B.


See https://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions
and https://en.wikipedia.org/wiki/Del#Directional_derivative

So it is all just an application of vector calculus.

 

[adelmakram]

The first part is follows from the chain rule: substituting B(x'+vt) for B':

∂B'/∂t = ∂B(x'+vt)/∂t = ∇B(x') ° ∂vt/∂t = ∇B(x') ° v.

But this can also be expressed as (v°∇)B.

But i have 3 concerns:
1) The above equation should equal to ∇B(x') ∂vt/∂t not to the composite ∇B(x') ° ∂vt/∂t
2) How (v°∇)B is driven from ∇B(x') ° v?
3) In wikipedia, it is a dot product (v.∇) B not a composite function (v°∇)B

 

[adelmakram]

∂B`(x`)/∂t = ∂B(x`+vt)/∂t

∂B/∂t= (∂B/∂x).(∂x/∂t) given that x=x`+vt

but ∂x/∂t= v

so ∂B/∂t= (∂B/∂x). v

 

[UltrafastPED]

Yeah, a dot not a composition. Couldn't find the dot in the menu. Gotta learn LaTex someday I keep saying to myself!

 

[adelmakram]

Yeah, a dot not a composition. Couldn't find the dot in the menu. Gotta learn LaTex someday I keep saying to myself!

Fine, so again how (v.∇) B is reached? it should be v (∇.B). In other words, the only operator that is acting on B should be ∇ not (v.∇).