# Vector Identity problem

1. Jun 5, 2013

### Krikri

1. The problem statement, all variables and given/known data
I want to compute the electric field knowing the magnetic field using a vector identity

2. Relevant equations

E=i $\frac{c}{k}$ (∇$\times$B)

B(r,t)=(μ0ωk/4π) ($\hat{r}$×$\vec{p}$)[1-$\frac{1}{ikr}$](eikr/r)

$\vec{p}$=dipole moment,constant vector

we have ti use the identity $\nabla$$\times$(A$\times$B)=(B$\cdot$∇)A-(A$\cdot$∇)B +A(∇$\cdot$B) +B(∇$\cdot$A)

the identy simplifies in this situtation because for some reason we take (A$\cdot$∇)B=0 and A(∇$\cdot$B)=0
So applying this we have :

E(r,t)=ic/k(μ0ωk/4π) $\nabla$[eikr/r2(1-$\frac{1}{ikr}$]×(r×p)+ic/k(μ0ωk/4π)[eikr/r2(1-$\frac{1}{ikr}$]∇×(r×p)
E(r,t)=i(ω/4πε0c)[ik($\frac{1}{r^2}$-$\frac{1}{ikr^3}$)]eikr r×(r×p) + i(ω/4πε0c)[(eikr/r^2)(1-$\frac{1}{ikr}$)][-∇$\cdot$r)p+(p$\cdot$∇)r] the this part says it's equal to -∇$\cdot$r)p+(p$\cdot$∇)r=-3p+p=-2p so

E(r,t)=$\frac{k^2}{4πε0}$(r×p)×r (ei(kr-ωt)/r) + $\frac{1}{4πε0}$[3r(r$\cdot$p)-p]($\frac{1}{r^3}$-$\frac{ik}{r^2}$)ei(kr-ωt)

My problem is i dont know how the vector identy is used here..with this tools we calculate magnetic and electric fields in the approximation zones( near,far-field) when vector potential is given. Can someone give a more simple example than this of what he did in this solution?

Last edited: Jun 5, 2013
2. Jun 5, 2013

### Simon Bridge

Lets make sure I follow you first:

You want to find $\vec{E}$ given:
$$\vec{E} = i\frac{c}{k}\vec{\nabla}\times\vec{B}\\ \vec{B}(\vec{r},t)=\frac{\mu_0\omega k}{4\pi}\left (\vec{r}\times\vec{p} \right ) \left [ 1 - \frac{1}{ik\vec{r}} \right ]\frac{1}{\vec{r}}e^{ikr}\\ \vec{\nabla}\times(\vec{A}\times\vec{B})=(\vec B\cdot\vec{\nabla})\vec A-(\vec A\cdot\vec \nabla)\vec B + \vec A(\vec \nabla \cdot \vec B)+ \vec B(\vec \nabla \cdot \vec A)$$... skipping a bit for now:
... if I got the above right, it looks to me that when you do $\vec \nabla \times \vec B$ you will end up with a term involving $\vec \nabla \times (\vec{r}\times\vec{p})$ ... which is where the identity should have come in.

BTW: the equation editor can be tricky to use.
It is normally better just to type the LaTeX markup in directly.

3. Jun 5, 2013

### Krikri

i figured out how the identity works. In this situation i don't know , but as it seems the problems i am into, don't require all of the above but simpler cases.

Thanks a lot for your time