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Perpendicular electric and magnetic field creates light or other electromagnetic waves. Is it a necessary property to have a perpendicular fields? If not what would happen when the fields are not perpendicular?

Timaeus
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Self-Made Man
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  • You may want to read up on Maxwell's equations that describe electromagnetic waves. That E (electric-field) is perpendicular to B (magnetic-field) is just the case in vacuum. It does not have to be true otherwise. – seb Apr 14 '13 at 17:18
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    A moving electric field, creates the perpendicular magnetic field. – Optionparty Apr 14 '13 at 19:36
  • Is there anything amphibian between electric and magnetic field? @Optionparty – Self-Made Man Apr 15 '13 at 15:31
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    @KaziarafatAhmed: Yes there is! In wave guides there modes of propagation. What we know as transverse electromagnetic field is just the TEM mode of a field. There are also the TE(transverse electric mode) with $B_z\neq 0$ and the TM(transverse Magnetic mode) with $E_Z\neq 0$ where $z$ is the direction of propagation! – Thanos Apr 18 '13 at 12:33
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    Just put a charge at rest in a magnetic field. The Coulomb field goes out in all directions, and will cross the magnetic field at every possible angle. Nothing special happens. – Michael Apr 18 '13 at 12:44
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    please note that if $\vec{E} \vec{B}$ is not equal to $0$, i.e. fields are not perpendicular, then there always exist a reference frame where $\vec{E}$ is parallel to $\vec{B}$. In particular, this is the case of neutrino field. – Murod Abdukhakimov Apr 18 '13 at 14:54
  • @Optionparty: A moving electric field, creates the perpendicular magnetic field. No. A changing (not "moving") electric field creates a magnetic field, which may or may not be perpendicular to it. In SciFi's Alien space ships were powered by crossing magnetic fields. Discussion of fictional physics is off-topic for this site. –  Aug 17 '13 at 18:27
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    @MurodAbdukhakimov: if ... fields are not perpendicular, then there always exist a reference frame where E is parallel to B Not true. The quantity $E\cdot B$ is a relativistic invariant. See http://en.wikipedia.org/wiki/Classification_of_electromagnetic_fields#Invariants –  Aug 17 '13 at 18:29
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    @Thanos: I think that should be an answer. –  Aug 17 '13 at 18:32
  • @Optionparty: You don't seem to be distinguishing between the concepts of electric charge and electric field. You might want to see if you can formulate a separate question that would help you to clatify your understanding of that issue. –  Aug 18 '13 at 16:31
  • @BenCrowell: You really think so? I'll just post it. Thank's! – Thanos Aug 19 '13 at 13:25
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    @BenCrowell Quote from page cited by you: "if $E\cdot B$ is non-zero there exists an inertial frame in which electric and magnetic fields are proportional." "Proportional" means "Parallel". – Murod Abdukhakimov Aug 22 '13 at 13:11
  • @MurodAbdukhakimov: Oops, sorry, you're right. –  Aug 22 '13 at 13:40
  • Only in percent of sources (currents or charges). In free space - never – MsTais Aug 03 '17 at 20:39

5 Answers5

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Yes there is! In wave guides, there are modes of propagation. What we know as transverse electromagnetic field is just the $TEM$ mode of a field. There are also the $TE$ (transverse electric mode) with $E_z≠0$ and the $TM$ (transverse Magnetic mode) with $B_z≠0$ where $z$ is the direction of propagation!

Aiden Cullo
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Thanos
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    But there you are talking about the multipole expansion of the EM-field (in Classical ED it might be not very explicit, but it is there). The fields still stay perpendicular to each other locally. Longitudinal components come from working with a spherical wavefront in cartesian basis, from what I understand. If this is not convincing, take Maxwell's equations in free space, go to reciprocal space and you will see that you NEVER can have a longitudinal component in absence of sources. – MsTais Aug 03 '17 at 20:36
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1) It is necessary for a plane EM wave. If one assumes solutions to the Maxwell's equation to be plane waves, it is not hard to show that $\vec B \cdot \vec E = 0$.

Namely, take the third Maxwell's equation and dot both sides with $\vec E$.

$$\nabla \times \vec E = - {{\partial \vec B} \over {\partial t}}$$ $$i\vec k \times \vec E = i\omega \vec B{\rm{ }}$$ $$(\vec k \times \vec E) \cdot \vec E = \omega \vec B \cdot \vec E$$ $$0 = \vec B \cdot \vec E$$ 2) In general case, it is not necessary.

Assume a constant current flowing up along the z axis. It produces static magnetic field around it in the form of concentric circles (and no electric field as net charge inside the conductor is zero).

Put a positive charge to a point (-1,0,0). It produces static electric field with radially outward lines.

At the point (0,1,0) the angle between electric and magnetic fields is 45 degrees.

enter image description here

Sergei Gorbikov
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Electricity can be static, like the energy that can make your hair stand on end. Magnetism can also be static, as it is in a refrigerator magnet. A changing magnetic field will induce a changing electric field and vice-versa—the two are linked. These changing fields form electromagnetic waves. Electromagnetic waves are formed by the vibrations of electric and magnetic fields. These fields are perpendicular to one another in the direction the wave is travelling. Once formed, this energy travels at the speed of light until further interaction with matter.

Maxwell developed a scientific theory to explain electromagnetic waves. He noticed that electrical fields and magnetic fields can couple together to form electromagnetic waves. The terms light, electromagnetic waves, and radiation all refer to the same physical phenomenon: electromagnetic energy. This energy can be described by frequency, wavelength, or energy. All three are related mathematically such that if you know one, you can calculate the other two. There’s a particle model and a wave model for electromagnetic radiation, and as we know electromagnetic radiation have dual nature. For an electromagnetic radiation to persist the two electric and magnetic fields have to propagate in a perpendicular direction to each other, EM waves consist of oscillating electric and magnetic fields.

hariom saranam
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    This doesn't really address the question of whether or not E and B fields can be non-perpendicular. It only states that the fields can only be perpendicular for EM radiation, which is misleading because there are situations in which you can have E and B fields that are not perpendicular – Jim Sep 15 '14 at 18:09
  • I don't think you read my question carefully. @hariom – Self-Made Man Sep 15 '14 at 18:17
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Electric and magnetic fields are always perpendicular to each other

user360682
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According to my understanding at physics, an electric field which is static cannot create any magnetic field.According to faraday for a magnetic field to exist there should be a varying electric filed(which varies over time).A simple example is battery connected to coil which is moved in and out near the galvanometer shows deflection(A galvanometer is device which detects magnetic filed by coupling magnetic filed near its vicinity to the coil inside and it internally has magnet which orients along the direction of earths's.The needle which is connected the coil coupling the magnetic field experiences a torque due to the EMF induced in it,and rests in the direction such that the earths magnetic field and the coupled magnetic are perpendicular to each other..hence the deflection of the galvanometer needle shows the stregth of magnetic field.

bobot
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    According to faraday for a magnetic field to exist there should be a varying electric filed(which varies over time). No. This is only one of the two source terms for magnetic fields in Maxwell's equations. –  Aug 17 '13 at 18:31