# [Physics] Electromagnetic radiation from AC circuit

electric-current

Is it possible that an AC circuit emit electromagnetic radiation?

I have read somewhere that due to change in polarity of the current continuously, electron oscillates in SHM at a mean position. During this situation can the electrons emit electromagnetic radiation?

All AC electric circuits radiate. Indeed, in the United States of America, all consumer electronic items must pass explicit FCC Electromagnetic Interference and FCC Radiation Safety assessment for their sale to be legal, so that one can make sure that the electromagnetic radiations emitted do not interfere with other electronic equipment and are safe for humans to be near. Most countries have similar regulations.

A lone electron (or any charge) oscillating in simple harmonic motion always radiates, as you describe. You can explicitly calculate this radiation from the Liénard–Wiechert potential for an accelerating charge. The Larmor Formula is a nonrelativistic approximation to this calculation.

You can, as a rough model, think of your circuit as a magnetic dipole loop antenna. Generally, radiation is very small unless the frequency is such that the circuit dimensions are comparable with a wavelength. A loop with radius $a\ll \lambda$ and carrying a sinusoidally varying current $I_0\,e^{i\,(k\,r-\omega\,t)}$ has farfield components given by:

$$E_\phi = \sqrt{\frac{\mu_0}{\epsilon_0}}\,\frac{k^2\,a^2\,\sin\theta}{4\,r}I_0\,e^{i\,(k\,r-\omega\,t)}$$ $$H_\theta= -\sqrt{\frac{\epsilon_0}{\mu_0}}\,E_\phi$$

where $k = 2\pi/\lambda$ is the wavenumber, $\lambda$ the wavelength, $\omega = k\,c$ the angular frequency and the fields are measured in a spherical co-ordinate system where $\theta$ measures the angle between the unit normal to the loop. You should be able to work out the total power radiated as:

$$P = \frac{\pi}{6}\,\sqrt{\frac{\mu_0}{\epsilon_0}}\,k^4\,a^4\,I_0^2$$

and estimate the loss from your circuit as:

$$P = \frac{1}{6\,\pi}\,\sqrt{\frac{\mu_0}{\epsilon_0}}\,k^4\,A^2\,I_0^2$$

where $A$ is the area of your circuit. You should be able to see that this is very small for $a\ll\lambda$.