# How do you tell whether charges oscillate in the antenna because of an electric or magnetic field?

The electrons in a receiving antenna oscillate, can we establish if they respond to an electric or a magnetic field?

How can we know if there is an electric field apart from the one caused by the magnetic field that makes the charges oscillate?

• a magnetic field is an electric field surely? someone please correct me if im wrong – Alex Robinson Apr 12 '17 at 8:21
• From Coulomb law $\vec{E}(x,y,z) = \frac{q \vec{r}}{4 \pi \epsilon_0 r^3},$ knowing that $c = \frac{1}{\sqrt{\epsilon \mu}}$ and using and using Biot-Savart law we get $\vec{B}(x,y,z) = \frac{\mu_0 e (\vec{v} \times \vec{r}) }{4 \pi r^3} = \frac{(\vec{v} \times \vec{r}) q}{c^2 (4 \pi \epsilon_0 r^3)} = \frac{\vec{v} \times \vec{E}}{c^2}$, ($r = \sqrt{x^2 + y^2 +z^2}$). From this follows that the magnetic field is an "electric" field from a moving charge. That's why in special relativity we put B & E into a symmetric tensor $F^{\mu\nu}$.And it transforms depending on who is observing – Mihai B. Apr 12 '17 at 8:59
• But only B component will for example induce an electric field in a coil. You could make a device such that only one component gets to activate the device. A device will respond differently to a B field than to an E field. It's all in Maxwell's equations. – Mihai B. Apr 12 '17 at 9:02
• A changing magnetic field may be inducing an electric field (and vice versa) but the Lorentz force on an electron is $$\vec{F} = q\vec{E} + {q\over c}(\vec{v} \times \vec{B}),$$ so the magnetic part of the force is suppressed by a factor of $v/c$ relative to the electric part - that's at least there orders of magnitude smaller for conduction electrons. To prevent misunderstanding, this is the Lorentz force in CGS units. Using SI units shifts the $c$ from the force law to the magnetic field, but of course the ratio of the forces does not change. – NickD Apr 15 '17 at 18:27