1
$\begingroup$

I came across the following question in a book I was studying:

Fmagnetic=μ0(M∇)H

Is this the correct expansion below? (I'm not too experienced with vectors operating on the gradient operator)

Fmagnetic=μ0[(Mx ∂H/∂x)i + (My ∂H/∂y)j + (Mz ∂H/∂z)k]

1 Answers 1

0

Assuming your M is a function of x,y,z (in vector form you have M = (Mx,My,Mz) where Mx,My,Mz map R^3 to R for each component) then

del(M) = (d/dx . Mx + d/dy . My + d/dz . Mz)H (I'm assuming everything is Cartesian not a general tensor) = (dMx/dx + dMy/dy + dMz/dz) H.

Now this will give you the product of two functions but if H is a vector (like M with each component have some transformation from R^3 -> R) then this means you use the scalar form of a*v = (a*v1,a*v2,a*v3) which means if H = (Hx,Hy,Hz) then the whole thing is equal to

(dMx/dx + dMy/dy + dMz/dz) *

Now M . grad(Hx) = . = Mx*dHx/dx + My.dHx/dy + Mz.dHx/dz

So they both look the same when they are expanded out, so I imagine you are right in your assertion. (It's been a while since I've done this kind of thing myself).

If I've made a mistake please let me know!