What happens if we use an antenna which is not a straight wire, i.e a loop or a coil? do we get EMR?

Now, if we do, what is the difference between this loop and the primary winding of a transformer? Why do we get radiowaves here and just a MF in the transformer?


1 Answer 1


A time varying loop certainly emits radiation - anything that accelerates nett charges or causes time varying departures from electro/magnetostatic conditions does - and this kind of antenna is known as the magnetic dipole antenna.

In principle, there is no difference between a loop antenna and a transformer primary, and the transformer primary will indeed radiate. A practical issue, though, is that one designs transformers so that the area of the loop is much smaller than the wavelength squared. This means that the power radiated is small enough not to be an issue in the application at hand. You can easily estimate the size of the effect by looking up the radiation resistance for a magnetic dipole antenna. From [1], §39.3.3 one finds that the power radiated from the loop is:

$$P = \mathcal{Z}_0\,\frac{\mu^2\,k^4}{6\,\pi}$$

where $k=\frac{2\,\pi}{\lambda}$ is the wavenumber, $\mu = n\,I\, A$ the RMS magnetic dipole moment of the loop, $I$ the RMS current, $A$ the loop's area, $n$ the number of turns and $\mathcal{Z}_0$ the freespace wave impedance, equal to $\sqrt{\frac{\mu_0}{\epsilon_0}}$ or about $377\Omega$. One often expresses this quantity as the radiation resistance:

$$R_{rad} = \frac{8\,\pi^3}{3}\,\mathcal{Z}_0\,\frac{n^2\,A^2}{\lambda^4}$$

From these equations, you can see that the radiated power is proportional to:


which is a very small ratio for most typical transformers. Let's think of a power transformer with a square meter cross section at $50{\rm Hz}$ line frequency. In this case, the ratio is a staggering $8\times10^{-28}$ (you read that right: eight times ten to the minus twenty eight!) Even when there are thousands of turns, the power lost through radiation is utterly negligible.

Note, however, that this does not mean that transformers always cause negligible electromagnetic interference. The above expressions deal with the farfield and power lost in the absence of other, nearby conductors that the transformer can couple to. The nearfields, however, are much bigger than one might guess from the radiation resistance expressions and the transformer's currents can significantly couple with nearby conductors through Mutual Inductance which is a distinct phenomenon from radiation and indeed the mechanism underlying the transformer's working in the first place.


[1] Lorrain & Corson, "Electromagnetic Fields and Waves", Third Edition, 1988

  • $\begingroup$ A uniform current circulating around a wire does not produce electromagnetic radiation, despite your first sentence. $\endgroup$
    – ProfRob
    Apr 19, 2017 at 7:45
  • $\begingroup$ The power formula is presumably also only valid for a sinusoidal AC current? The general form would involve the time average of the second derivative of the magnetic dipole moment. $\endgroup$
    – ProfRob
    Apr 19, 2017 at 7:50
  • $\begingroup$ @RobJeffries You're right and the formulas reflect that - see revised first line. However, let's think some more: why, then, was is a problem with the Bohr atom, for instance? In that case, you have a moving nett charge, imperfectly shielded by the proton. This raises an interesting point: in a wholly classical situation, you would also have radiation from wires as the electrons moved between the ions. However, what would QED / delocalized wavefunction say about electrons moving in a periodic lattice? second point: Evidently the formulas are for time-sinusoidal; the general case is almost .. $\endgroup$ Apr 19, 2017 at 10:35
  • $\begingroup$ .... as you say; the farfields vary as the second derivative of the magnetic moment, and their time averaged inner product (i.e. flux of Poynting vector over the big sphere) is the mean square second derivative, as in the farfield terms in Feynman's formula from the Liénard-Weichert potentials. $\endgroup$ Apr 19, 2017 at 10:38
  • $\begingroup$ The problem is with a uniform current. If the charges are single, or asymmetrically distributed, and going around in a circle then you do get a net radiation field. $\endgroup$
    – ProfRob
    Apr 19, 2017 at 11:30

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