I don't quite understand why the force is perpendicular, can somebody explain with a magnetic field diagram as to how the perpendicular direction is produced as a result of the interaction of the two fields (one of the wire and the other we apply).

I know the force is proportional to the cross product of the length and the field and hence the perpendicular direction, but how do the individual fields interact to produce the force.

  • $\begingroup$ It is enough that the current carrying wire and the external field are not parallel. The dot product expresses the area between the current and the magnetic field directions and strengths. It is maximum for perpendicular vectors and zero for parallel. $\endgroup$ – HolgerFiedler Aug 9 '16 at 14:56
  • $\begingroup$ That the resulting force is perpendicular to both vectors has to do with the gyroscopic effect of the involved electrons with their magnetic dipole moments and their intrinsic spin. $\endgroup$ – HolgerFiedler Aug 9 '16 at 14:59
  • $\begingroup$ See About the Lorentz force $\endgroup$ – HolgerFiedler Aug 9 '16 at 15:01
  • $\begingroup$ @Batwayne what fields would be interacting? I could assume the magnetic field generated by the current in the wire, and the magnetic field the wire is passing through. But I want to be sure. Either way the overlapping magnetic fields of the wire and the magnetic field can be summed, and would produce an effect at a higher order approximation of the system. But the effect of the overlapping magnetic fields would be negligible, and won't be enlightening as to why the force is perpendicular to both the magnetic field and the velocity of the charged particle. See my answer below. $\endgroup$ – snarkyname77 Aug 9 '16 at 15:52

Why is the force perpendicular? It may be helpful to reverse the question. What would happened if the force where parallel or antiparallel to the velocity of the particle? If the force where parallel or antiparallel to the particle's velocity the particle would accelerate along its direction of travel. Either increasing or decreasing in velocity.

Applying the constraint of conservation of energy to the system. If the velocity of the charged particle where to increase. Where would the energy for the increase in velocity come from? If the velocity of the particle decreased. Where would the energy lost by the particle go? There are only two elements to the system, the magnetic field, and the charged particle. If the energy in the particle increases then the field's energy must decrease. If the energy of the particle decreases then the field's energy must increase. How would the energy be transferred?

The force being perpendicular to the magnetic field and the motion of the particle observes conservation of energy. While the velocity of the particle changes the energy of the particle is conserved.

  • $\begingroup$ Okay, fair point, but why not just make the magnetic field point inward to the charge that generated it like the electric and gravitational fields? Then a charge wouldn't go forever along a field line because it would hit the source and stop. The reason magnetic fields do all the funny cross product perpendicular stuff is because of how they're linked to electric fields through special relativity. $\endgroup$ – EL_DON Aug 9 '16 at 19:32
  • $\begingroup$ @EL_DON As for the direction of the magnetic field lines. Observation dictates whether a hypothesis becomes theory. Theory does not dictate observation. So magnetic field lines pointing in towards a current carrying wire, like the electric and magnetic field, is a conjecture that does not yield a hypotheses that matches observation. There is nothing stopping you or anyone else from developing a hypothesis around the conjecture that magnetic field lines point radially inward or outward from a current carrying wire. But that hypothesis would not match observation and would not be useful. $\endgroup$ – snarkyname77 Aug 10 '16 at 15:13
  • $\begingroup$ @EL_DON which magnetic field are you referring too? The external magnetic field that the charged particle(s) move through, or the magnetic field generated by moving charged particle(s)? $\endgroup$ – snarkyname77 Aug 10 '16 at 16:06
  • $\begingroup$ @EL_DON finally your last statement "The reason magnetic fields do all the funny cross product perpendicular stuff is because of how they're linked to the electric fields through special relativity." Again, theory does not dictate observation. To say that the 'reason' for the magnetic field directionality is because of the electromagnetic field tensor is misleading. $\endgroup$ – snarkyname77 Aug 10 '16 at 16:15
  • $\begingroup$ Observation gives the what and theory gives the why. So yes, it is correct to use a theory to explain the reason why observations are the way they are. $\endgroup$ – EL_DON Aug 10 '16 at 19:48

The magnetic field is the Lorentz transform of the electric field. If we boost to the reference frame of the moving charges in your current, an electric field will emerge. The resulting electric force is perpendicular to both the magnetic field and the velocity of the charges in the current. Let's see an example:

There is a test charge $q$ moving with velocity $v$ to the right in the lab reference frame. A short distance below it is a linear charge distribution with linear charge density $\lambda$. Superimposed on this is another linear charge distribution with linear charge $-\lambda$, which is moving with velocity $u$ to the right. Because $\lambda-\lambda=0$, there appears to be no electric field in the lab frame. Now boost to velocity $v$ to the right so that the test charge $q$ is at rest. The positive line charge experiences relativistic length contraction such that the distance between charges on the wire shorter and the charge density is higher. The new positive charge density is now $\gamma \lambda$, where $\gamma$ is the relativistic time dialation / length contraction factor. The negative line charge, however, was already moving to the right, and was already length contracted. By boosting, we undo some of the length contraction and therefore reduce the linear charge density of the negative group of charges. Now the density of positive charges does not equal the density of negative charges anymore and the net charge is no longer zero! An electric field emerges! The electric force on the test charge, which is stationary in this reference frame, is $F=qE$, and this force is perpendicular to the boost velocity (which is the test charge velocity in the lab frame) and to the magnetic field that we saw in the lab frame.

There are still magnetic fields in the boosted frame, but they are different and a piece has been converted into an electric field.

  • $\begingroup$ I don't see how the perpendicular direction of the magnetic force emerges in this description. You seem to skip over that point in the last sentence of your main paragraph. $\endgroup$ – sammy gerbil Aug 9 '16 at 20:26
  • $\begingroup$ @sammy gerbil: in one frame there is a net electric charge and it makes a force away from the wire. The relativistic boost is perpendicular to this force, so the amount of force is the same in both frames. One frame does not have a net charge, so theRe is no $E$ field to make the perpendicular force, so the magnetic force must be in this direction. As for the field itself, this just comes out of the mathematics of the Lorentz transform. $\endgroup$ – EL_DON Aug 9 '16 at 23:52

That the force which a magnetic field exerts on a moving charge is perpendicular to both the applied magnetic field and the velocity of the charge, is a fact of observation from which - along with other observational facts - the theory of electromagnetism was derived. The equation for the Lorentz force $\vec F=q\vec v \times \vec B$ is an expression of this observation.

The magnetic field generated by a moving charge in the wire does not act on itself. So it is not the interaction (or superposition) of the magnetic fields which produces this force. However, this charge-generated magnetic field can exert a force on another moving charge in another part of the same wire if it is bent around, eg into a loop.

  • $\begingroup$ True, but doesn't explain why. $\endgroup$ – EL_DON Aug 9 '16 at 19:33

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