Register to reply 
Find the power radiated using the Poyting vector 
Share this thread: 
#1
Dec2111, 08:53 AM

P: 241

1. The problem statement, all variables and given/known data
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 zaxis. 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] (tr/c)] A=(mu(subscript 0) p(subscript 0) omega^2 d)/(4pi cr) cos [itex]\theta[/itex] cos [[itex]\omega[/itex] (tr/c)] zhat 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.nhat da 3. 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. 


#2
Dec2111, 02:15 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,690

You want to integrate over a closed surface, e.g. a sphere, to find the total power being radiated away. This is just like when you used the integral form of Gauss's law except instead of integrating the electric field, you're integrating the Poynting vector.



#3
Dec2411, 06:38 AM

P: 241

I found an easier way to do this. From my notes it says:
P=[itex]\oint_{S}[/itex] N.nhat da =(1/μ[itex]_{0}[/itex]) [itex]\int^{2pi}_{0}[/itex][itex]\int^{pi}_{0}[/itex]B[itex]_{\phi}[/itex]E[itex]_{\theta}[/itex] r[itex]^{2}[/itex] sin[itex]\theta[/itex] d[itex]\theta[/itex] d[itex]\phi[/itex] =(1/μ[itex]_{0}[/itex]) [itex]\int^{2pi}_{0}[/itex][itex]\int^{pi}_{0}[/itex] [itex]\frac{\mu_{0}I_{0}\delta l}{4\pi}[/itex] sin[itex]\theta[/itex] ([itex]\frac{i\omega}{rc}[/itex]+[itex]\frac{1}{r^{2}}[/itex]exp(i(kr  omega*t)) [itex]\frac{\mu_{0}I_{0}\delta l}{4\pi}[/itex] sin[itex]\theta[/itex] sin [itex]\theta[/itex] (i*omega/r) exp(i(kr  omega*t)) sin [itex]\theta[/itex] d[itex]\theta[/itex] d[itex]\phi[/itex] One problem: what do I do with the [itex]\delta[/itex]l's? How are they to be gotten rid of? Do I just ignore them? 


#4
Dec2411, 08:23 AM

P: 263

Find the power radiated using the Poyting vector
There is no dl at all. You didnt plug the proper fields in the calculation of the poynting vector. You should be calculating electric quadrupole radiation not magnetic dipole radiation. The fields you wrote for the poynting vector are the fields of a magnetic dipole.
Write the fields which u calculated for the electric quadrupole.. 


#5
Dec2411, 09:09 AM

P: 241

"Write the fields which u calculated for the electric quadrupole.."
I don't know how to do that. How do I go from a dipole field to a quadrupole field then? Multiply everything by 2? That doesn't seem right. Yes, I will look in my notes and books and see if they say anything, but I am asking just in case this turns out to be one of those complicated questions that require 'thinking' rather than bookwork. 


#6
Dec2411, 09:11 AM

P: 241

Never mind, I just realised I copied down the equations for B and E from the question above the question I was supposed to be doing.



#7
Dec2411, 09:16 AM

P: 263

no, the quadrupole is the next term in the multipole expansion of the fields. After the dipole comes the quadrupole in the expansion.



#8
Dec2411, 10:53 AM

P: 241

So
P=[itex]\oint[/itex]N.nhat da =[itex]\int[/itex] div N d(tau) =[itex]\int[/itex] [itex]\frac{\partial N_{z}}{\partial z}[/itex] d(tau) but N has no z component so the integral goes to zero. That the power seems zero doesn't seem right. 


#9
Dec2411, 10:57 AM

P: 241

I sort of ignored the nhat because I didn't see any other way to go about doing the question. What would be the direction of nhat anyway? All we know about the quadrupole by considering two dipoles is:
+p[itex]_{0}[/itex] cos ωt at z=d/2 p[itex]_{0}[/itex] cos ωt at z=d/2 so nhat could be pointing in either the rhat or thetahat direction. Does the nhat need to be taken into account to do the question properly? 


#10
Dec2411, 11:13 AM

P: 241

Previously I made a mistake while posting.
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)))] zhat I had missed the zhat out at the end. 


#11
Dec2811, 03:30 PM

P: 241

I changed it from cylindrical to spherical coordinates. I found that in spherical coordinates it only has a rhocomponent. So only [itex]\frac{1}{\rho^{2}}[/itex] [itex]\frac{\partial}{\partial\rho}[/itex] ([itex]\rho^{2}[/itex] N[itex]_{\rho}[/itex]) matters when trying to work out div N.
But N[itex]_{\rho}[/itex] has no [itex]\rho[/itex] terms in it. So div N=0. This is the same as the result I got when I did it in cylindrical coordinates. I spoke to someone else who is doing the same homework and he also doesn't think it makes sense for the power radiated to be zero. Is the power radiated really zero? If so, why? If not, is there some sort of trick that will help one to answer the question? Thanks if anyone replies. 


#12
Dec2911, 03:13 AM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,690

What did you get for E and B?



#13
Dec2911, 06:19 AM

P: 241

E=grad[itex]\phi[/itex]
=[itex]\frac{\partial\phi}{\partial r}[/itex]rhat+[itex]\frac{1}{r}[/itex][itex]\frac{\partial\phi}{\partial\theta}[/itex]θhat +[itex]\frac{\partial\phi}{\partial z}[/itex]zhat =[[itex]\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi}[/itex][itex]\frac{cos^{2}θ}{r}[/itex]([itex]\frac{\omega}{c}[/itex]sin([itex]\omega[/itex](t[itex]\frac{r}{c}[/itex]))+[itex]\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi}[/itex](2sinθ [itex]\frac{cosθ}{r}[/itex])cos([itex]\omega[/itex](t[itex]\frac{r}{c}[/itex]))]rhat+[[itex]\frac{\mu_{0}p_{0}\omega^{2}d}{4r^{2}\pi}[/itex]sinθ cosθ cos(ω(t[itex]\frac{r}{c}[/itex]))]θhat =[itex]\frac{\mu_{0}p_{0}\omega^{2}d cosθ}{4\pi r}(cosθ \frac{\omega}{c}sin(\omega[/itex](t[itex]\frac{r}{c}[/itex])))rhat+([itex]\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r^{2}}[/itex]sinθcosθcos([itex]\omega(t\frac{r}{c}[/itex])))θhat B=[itex]\nabla[/itex][itex]\times[/itex]A =[itex]\frac{1}{r}[/itex][itex]\frac{\partial A_{z}}{\partialθ}[/itex]rhat [itex]\frac{\partial A_{z}}{\partial r}[/itex]thetahat =[itex]\frac{\mu_{0}p_{0}\omega^{2}d sinθ}{4\pi cr^{2}}[/itex]sinθcos(ω(tr/c)))rhat+([itex]\frac{\mu_{0}p_{0}\omega^{2}d sinθ}{4\pi cr}[/itex]cosθ([itex]\frac{ω}{c}[/itex]sin(ω(tr/c))[itex]\frac{sinθ}{r}[/itex]cos(ω(tr/c)))θhat Have I gone wrong somewhere? 


#14
Dec2911, 12:22 PM

P: 241

Sorry, I think I know where I have gone wrong now. E is not grad phi in this case.



#15
Dec2911, 12:30 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,690

What were your results in spherical coordinates? You probably don't want to use cylindrical coordinates here.
Even if you do, you have to remember to rewrite the potentials in terms of the cylindrical coordinate ##(r, \theta, z)## first because ##r## and ##\theta## aren't defined the same way in cylindrical and spherical coordinates. Since you're looking for the fields in the radiation zone, where r is large, you can drop contributions proportional to 1/r^{2}. The 1/r terms will dominate. 


#16
Dec2911, 12:54 PM

P: 241

So I have to find E and B in spherical coordinates? I can't just find them in cylindrical coordinates, and then convert them to spherical when I find N? Because as I said, when I convert the result I got for N into spherical coordinates, and then I do the integral, it just goes to zero.
I have realised more now that the question is about a quadrupole, but it is asking you to consider two dipoles. Does that mean I have to multiply everything I did by two? What is a 'radiation zone'? Thanks if anyone explains. 


#17
Dec2911, 01:04 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,690

Despite the presence of ##\hat{z}##, the expressions for the potentials are, I'm pretty sure, in spherical coordinates. (You should check that.) So using spherical coordinates is the path of least resistance. Also, the end result won't and, more important, can't matter on which coordinate system you choose.
What textbook are you using? For terms like radiation zone, you should look those up in your text rather than asking here. You want to learn how to find that kind of basic information on your own. 


#18
Dec2911, 01:42 PM

P: 241

I am using Griffiths. It does not explain what radiation zone is. It just highlights it in bold. It says 'we are interested in fields that survive at large distances from the source, in the so called radiation zone'.
'large distances'? That means the radiation zone is everywhere? If the radiation zone is everywhere, and there is no difference between one place and another with regards to whether it is in the radiation zone or not, then why is it mentioned in the question? 


Register to reply 
Related Discussions  
Total Power Radiated  Introductory Physics Homework  2  
Power radiated by the Sun  Introductory Physics Homework  11  
Power radiated by the sun  Introductory Physics Homework  0  
Radiated power, what does this even mean?  Introductory Physics Homework  9  
Net Power vs. Net Radiated  Introductory Physics Homework  1 