- #1
Krikri
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Homework Statement
I want to compute the electric field knowing the magnetic field using a vector identity
Homework Equations
E=i [itex]\frac{c}{k}[/itex] (∇[itex]\times[/itex]B)
B(r,t)=(μ0ωk/4π) ([itex]\hat{r}[/itex]×[itex]\vec{p}[/itex])[1-[itex]\frac{1}{ikr}[/itex]](eikr/r)
[itex]\vec{p}[/itex]=dipole moment,constant vector
we have ti use the identity [itex]\nabla[/itex][itex]\times[/itex](A[itex]\times[/itex]B)=(B[itex]\cdot[/itex]∇)A-(A[itex]\cdot[/itex]∇)B +A(∇[itex]\cdot[/itex]B) +B(∇[itex]\cdot[/itex]A)
the identy simplifies in this situtation because for some reason we take (A[itex]\cdot[/itex]∇)B=0 and A(∇[itex]\cdot[/itex]B)=0
So applying this we have :
E(r,t)=ic/k(μ0ωk/4π) [itex]\nabla[/itex][eikr/r2(1-[itex]\frac{1}{ikr}[/itex]]×(r×p)+ic/k(μ0ωk/4π)[eikr/r2(1-[itex]\frac{1}{ikr}[/itex]]∇×(r×p)
E(r,t)=i(ω/4πε0c)[ik([itex]\frac{1}{r^2}[/itex]-[itex]\frac{1}{ikr^3}[/itex])]eikr r×(r×p) + i(ω/4πε0c)[(eikr/r^2)(1-[itex]\frac{1}{ikr}[/itex])][-∇[itex]\cdot[/itex]r)p+(p[itex]\cdot[/itex]∇)r] the this part says it's equal to -∇[itex]\cdot[/itex]r)p+(p[itex]\cdot[/itex]∇)r=-3p+p=-2p so
E(r,t)=[itex]\frac{k^2}{4πε0}[/itex](r×p)×r (ei(kr-ωt)/r) + [itex]\frac{1}{4πε0}[/itex][3r(r[itex]\cdot[/itex]p)-p]([itex]\frac{1}{r^3}[/itex]-[itex]\frac{ik}{r^2}[/itex])ei(kr-ωt)
My problem is i don't 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?
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