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Homework Help: Hertzian dipole consistent with Biot Savart law at small distances

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A Hertzian dipole is located at the origin of spherical coordinates and is aligned with the θ=0 direction. The dipole has strength I(subscript 0)[itex]\delta[/itex]l and oscillates with angular frequency [itex]\omega[/itex]. The magnetic field that it produces is given by the real part of the expression:

    B(r)=[([itex]\mu[/itex](subscript 0) I(subscript 0) [itex]\delta[/itex] l)/(4pi)] sin[itex]\theta[/itex] [((-i [itex]\omega[/itex])/(rc))+(1/(r^2))]exp [i(kr-[itex]\omega[/itex]t)] phi-hat

    When grouping terms by r-dependence, there are essentially three contribution to the fields produced by a Hertzian dipole: the r^-3 terms are effectively the electrostatic field; the r^-2 terms give rise to what is called the induction field; and, the radiation field is the single term r^-1.

    i) Show that, in the limit of small distances r and zero angular frequency, the amplitude of the field given by this expression is consistent with the Biot-Savart law:

    B(r)=(([itex]\mu[/itex](subscript 0) I)/(4pi) (([itex]\delta[/itex] l cross r-hat)/(r^2))

    ii)Show also that in the limit of large r the form of the expression is consistent with that required for a radiation field.

    3. The attempt at a solution

    B(r)=[([itex]\mu[/itex](subscript 0) I(subscript 0) [itex]\delta[/itex] l)/(4pi)] sin[itex]\theta[/itex] [((1/(r^2))]exp [i(kr-[itex]\omega[/itex]t)] phi-hat

    B(r)=[([itex]\mu[/itex](subscript 0) I(subscript 0) [itex]\delta[/itex] l)/(4pi)] sin[itex]\theta[/itex] [((1/(r^2))] cos(kr-[itex]\omega[/itex]t) phi-hat

    B(r)=[([itex]\mu[/itex](subscript 0) I(subscript 0) [itex]\delta[/itex] l)/(4pi)] sin[itex]\theta[/itex] [((1/(r^2))] cos(kr) phi-hat

    B(r)=[([itex]\mu[/itex](subscript 0) I(subscript 0) [itex]\delta[/itex] l)/(4pi)] sin[itex]\theta[/itex] [((1/(r^2))] cos(0) phi-hat

    because r is small

    B(r)=[([itex]\mu[/itex](subscript 0) I(subscript 0) [itex]\delta[/itex] l)/(4pi)] sin[itex]\theta[/itex] [((1/(r^2))] phi-hat

    I suppose I need to find out what theta would be, but I am having trouble visualising the thing in my head. I suppose theta would have to be 90 degrees, but why?

    How does one go from delta l to delta l cross r-hat? And how is the phi-hat gotten rid of?

    Please help.
  2. jcsd
  3. Dec 23, 2011 #2
    And I have no idea how they went from there being an I(subscript 0) in the original expression, to it just being 'I' without the sucript in the Biot-Savart law. What does the I(subscript 0) represent anyway? The initial current? How can there be an initial current? It doesn't appear as if the current is changing. Please help.
  4. Dec 24, 2011 #3
    The problem is that the given quantity is not the magnetic field, but the magnetic potential vector.
    So you have to take the curl of this quantity to get the magnetic field....
  5. Dec 24, 2011 #4
    So B(r)=[(μ(subscript 0) I(subscript 0) δ l)/(4pi)] sinθ [((-i ω)/(rc))+(1/(r^2))]exp [i(kr-ωt)] phi-hat

    is the magnetic vector potential? So it is being suggested that the lecturer who set the homework made a mistake in defining the above equation as B?

    I think I will look up what the magnetic vector potential is. But I am going to ask now why you think the above equation represents the magnetic vector potential, just in case after I look it up I am still confused. I'd be grateful if you could explain please.
  6. Dec 24, 2011 #5
    The simplest i could say is check the units...

    The good way would of course be to derive that.
    Solve Maxwell's equations for the vector potential in the Lorentz gauge. You get an inhomogenous wave equation. Solve it with Fourier transforms. And you will get the vector potential from that. Then consider the multipole expansion of the vector potential. The second term will be the magnetic dipole you are seeking.

    And yes the lecturer indeed made a pretty bad mistake....
  7. Dec 24, 2011 #6
    And check : "J. D. Jackson: Classical electrodynamics (3rd edition), Chapter 9"
  8. Dec 25, 2011 #7
    Apparently, the dipole is aligned with theta = 0, and the expression gives B for all space, so the given equation for B is not wrong.

    But still, if theta=0, why does the whole expression for B just not go to zero if there is a sin theta in there? Why does it tnd to the Biot-Savart law instead?
  9. Dec 25, 2011 #8
    The given equation for B is wrong. It is not the equation for B. it is the equation for the vector potential...
    Take the curl of this and you get the equation for B.
  10. Dec 26, 2011 #9
    Thaakisfox, thank you for your help, but since you are the only person who said that the lecturer who set the problem sheet made a mistake, I am still going to assume that there was no mistake in order to try to do the question, because I really don't feel like taking the curl of something right now.

    I had a look at part ii) of the question.

    ii) Is the answer just that for large distances 1/(r^2)=0? But why would the radiation term remain? Why is complex term (-i*ω/rc) not taken into account for small distances, but taken into account for large distances? Would exp[i(kr-ω t)] be taken into account for large distances? If so, why, if not, why not? I noticed that the question is worth three marks, so I can't believe that all one needs to say that 1/(r^2)=0 for large distances. What else should be said?
  11. Dec 26, 2011 #10
    And for part i), also assuming that there was no mistake in the question:

    I realised that sin θ=θ for small distances, and θ [itex]\times[/itex] [itex]\phi[/itex]-hat=r-hat, but how would one go from θ to θ-hat? How can (θ [itex]\phi[/itex]-hat) be equivalent to r-hat? Can one just change the θ to θ-hat and then just take the cross product? If so, why? If not, why not? And how does [itex]\delta[/itex] l become [itex]\delta[/itex] l, because l was written in bold for the Biot-Savart law but it was not written in bold in the long expression above. And why is it delta l [itex]\times[/itex] r-hat? How does one get the cross product sign from the long expression?

    I'd appreciate it if anyone helps please.
  12. Dec 27, 2011 #11
    i see you are having some problems with basic vector manipulation. But that is not a problem at all, we are all here to learn :))
    Sooo the variable you need to consider is the distance r. The angle theta just defines that which point at distance r are we looking at. It is an angle variable. So you have to leave the sine as it is.
    Next point is that the cross product is a product of two vectors. Theta is not a vector. Is an angle. Theta hat is indeed a vector. It is the unit vector pointing towards the direction which theta grows.
    Now consider r to be small. Then u neglect the radiation field exp(ikr)/r wrt the other part.
    Next point is the vector dl. Vector dl is dl times phi hat. As it is pointing in the direction of growing phi. We model the magnetic dipole . dl is the infinitesimal wire length of through which the current generatimh the magnetic dipole moment is passing. Hence vector dl is equal to dl times phi hat.
    Hope that is clear till now. Lets see where you got to now.
    But do tell your lecturer that mixed the stuff up.
    What he did was substituted the magnetic dipole moment instead of the electric dipole moment in the field of electric dipole radiation. But that is not good. The magnetic dipole radiation is the next term in the series and the magnetic and electric fields are a bit different (they r exchanged compared to the electic dipole radiation.) so do tell him to correct his mistake.
  13. Jan 3, 2012 #12
    Using δl [itex]\phi[/itex]-hat=[itex]\delta[/itex]

    B(r)=[itex]\frac{\mu_{0}I_{0}θ}{4\pi r^{2}}[/itex]delta l

    But how do I turn the θ into [itex]\times[/itex]r-hat?

    Thanks if anyone replies.
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