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Find the power radiated using the Poynting vector
As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, +p[itex]_{0}[/itex] cos [itex]\omega[/itex] t at z=+d/2, and -p[itex]_{0}[/itex] cos [itex]\omega[/itex] t at z=-d/2, hence separated by a distance d along the z-axis. In the radiation zone, the scalar and vector potentials are given by:
[itex]\phi[/itex]=-([itex]\mu[/itex](subscript0) p(subscript 0) omega^2 d/4pi)((cos^2 theta)/r) cos [[itex]\omega[/itex] (t-r/c)]
A=-(mu(subscript 0) p(subscript 0) omega^2 d)/(4pi cr) cos [itex]\theta[/itex] cos [[itex]\omega[/itex] (t-r/c)] z-hat
a) Find the electric and magnetic fields, E and B.
b) Find the Poynting vector, N, and the power radiated, P= loop integral (subscript S) N.n-hat da
a) I have done this but I am not writing all of it out.
b) N= E cross H
= [([itex]\mu[/itex](subscript 0) p(subscript0)^2 [itex]\omega[/itex]^4 d^2)/([itex]\mu[/itex] *16 pi^2 c r^2)] cos^3 [itex]\theta[/itex] [[([itex]\omega[/itex]^2)/(c^2)] sin^2 ([itex]\omega[/itex](t-(r/c))) - ([itex]\omega[/itex]/c) (sin [itex]\theta[/itex])/([itex]\mu[/itex][itex]\mu_{0}[/itex] r) cos ([itex]\omega[/itex](t-(r/c)))
-2sin[itex]\theta[/itex] cos([itex]\omega[/itex](t-(r/c)))((p[itex]_{0}[/itex]/[itex]\mu[/itex]) [([itex]\omega[/itex]^2 d)/(4pi c r)] cos[itex]\theta[/itex](([itex]\omega[/itex]/c) sin([itex]\omega[/itex](t-(r/c)) - (sin[itex]\theta[/itex]/[itex]\mu[/itex][itex]\mu_{0}[/itex] r) cos([itex]\omega[/itex](t-(r/c)))
But how does one work out the power radiated from this? My books and notes don't provide a simple answer. I am having trouble visualising the electric quadrupole and don't know what the unit vector is. And it has to be integrated over the area. The area of what? How can there be an area when it is two electric dipoles? Please help.
Homework Statement
As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, +p[itex]_{0}[/itex] cos [itex]\omega[/itex] t at z=+d/2, and -p[itex]_{0}[/itex] cos [itex]\omega[/itex] t at z=-d/2, hence separated by a distance d along the z-axis. In the radiation zone, the scalar and vector potentials are given by:
[itex]\phi[/itex]=-([itex]\mu[/itex](subscript0) p(subscript 0) omega^2 d/4pi)((cos^2 theta)/r) cos [[itex]\omega[/itex] (t-r/c)]
A=-(mu(subscript 0) p(subscript 0) omega^2 d)/(4pi cr) cos [itex]\theta[/itex] cos [[itex]\omega[/itex] (t-r/c)] z-hat
a) Find the electric and magnetic fields, E and B.
b) Find the Poynting vector, N, and the power radiated, P= loop integral (subscript S) N.n-hat da
The Attempt at a Solution
a) I have done this but I am not writing all of it out.
b) N= E cross H
= [([itex]\mu[/itex](subscript 0) p(subscript0)^2 [itex]\omega[/itex]^4 d^2)/([itex]\mu[/itex] *16 pi^2 c r^2)] cos^3 [itex]\theta[/itex] [[([itex]\omega[/itex]^2)/(c^2)] sin^2 ([itex]\omega[/itex](t-(r/c))) - ([itex]\omega[/itex]/c) (sin [itex]\theta[/itex])/([itex]\mu[/itex][itex]\mu_{0}[/itex] r) cos ([itex]\omega[/itex](t-(r/c)))
-2sin[itex]\theta[/itex] cos([itex]\omega[/itex](t-(r/c)))((p[itex]_{0}[/itex]/[itex]\mu[/itex]) [([itex]\omega[/itex]^2 d)/(4pi c r)] cos[itex]\theta[/itex](([itex]\omega[/itex]/c) sin([itex]\omega[/itex](t-(r/c)) - (sin[itex]\theta[/itex]/[itex]\mu[/itex][itex]\mu_{0}[/itex] r) cos([itex]\omega[/itex](t-(r/c)))
But how does one work out the power radiated from this? My books and notes don't provide a simple answer. I am having trouble visualising the electric quadrupole and don't know what the unit vector is. And it has to be integrated over the area. The area of what? How can there be an area when it is two electric dipoles? Please help.
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