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According to Biot-Savart Law, if there is a charged particle in motion, there will be a magnetic field. My question is whether the counterpart of this law also holds true, i.e. if there is a magnetic field, whether there will be a charged particle be in motion. Let me explain my question a bit more clearly. First of all I don't know whether we can have a single charge in motion and another at rest in reality. If we can have then there may be two cases:

Case 1: At time t=0 a charged particle is placed at rest at point P where B=0 and at time t>0 an external B field is created in a region R covering the point P.

Case 2: At time t=0 a magnetic field B is created in a region R. At time t>0 a charged particle is placed at rest at point P in R.

Note of Caution: The point P is chosen in such a way that there won't be any external electric field at P.

Now come to the question: Will the charged particle move? If yes, under which case?

If 'NO', whether the charged particle will absorb any energy or not?

Subhra
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2 Answers2

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Magnetic fields can be generated by particles in motion (in the classical sense). Generally, this not a requirement. For example, magnetism can arise from quantum effects, such as orbital angular momentum (you could say that this is a particle in motion in the classical sense, but quantum probabilities come into play here) and may also arise due to the spin state of a particle (not linear motion in the traditional sense).

user3814483
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  • So you are looking magnetic field as the effect of motion of electric charge. I wish to know whether, the motion of electric charge can be viewed as the effect of magnetic field. – Subhra Jul 12 '14 at 08:02
  • I was distinguishing between macroscopic fields (due to particles in macroscopic motion), and fields that arise due to quantum effects (orbital angular momentum and spin). I generally don't call the latter "motion" as people typically take that to mean classical motion. So it was a technical exception.

    As to your comment, recall from the Lorentz force equation that magnetic fields cannot do work. Therefore, charges cannot be put into motion from rest (in a classical sense) simply because there is a magnetic field.

     – user3814483 Jul 12 '14 at 16:19
  • In Lorentz force law, the force is dependent on both the velocity of the charged particle as well as the applied magnetic field. So F=F(v,B) and F(v,0)=F(0,B)=F(0,0)=0 in this case. Can you please let me know whether v and B are independent of each other in Lorentz Force Equation? – Subhra Jul 12 '14 at 17:16
  • Your equations are correct. Initially, the v and B are independent (at t= 0, the particle starts of with some velocity v_0 in a field B, each set independently). Once the field deflects the particle due to the v_0 x B term, it will change the direction of v_0, let's call this v_1. At a slightly later instance in time, you would re-compute the deflection by evaluating v_1 x B. Remember, the Lorentz force (no electric fields) is just m*dv/dt = v x B. So the effect of v x B "feeds back" onto the v. – user3814483 Jul 12 '14 at 18:17
  • Your answer is really convincing. It seems we can formulate v=v(B) explicitly. Your last sentence is teasing me to think of the interaction of a charged particle in a magnetic field as a combination of Action and Reaction. Please let me know with explanation whether it is correct. – Subhra Jul 13 '14 at 04:39
  • Yes, in a sense, it is. I didn't include vector notation in my last post; generally we're dealing with a system of three equations and three unknowns (dv_x / dt, dv_x / dt, dv_z / dt). For this reason, for each scalar equation the component of v on the left side won't be the same as the one on the right side, but they will be coupled. Once B is known as a function of space and time, a system of coupled differential equations must be solved. Then you'll arrive at your explicit solution for all three velocity components. – user3814483 Jul 13 '14 at 05:02
  • Can you please explain this in terms of action and reaction, if it is so? – Subhra Jul 13 '14 at 05:21
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The only exceptions I know arise from quantum effects. For example a charged fundamental particle with non-zero spin also creates a magnetic field. So an isolated static electron has a non-zero magnetic moment. Even though this is a purely quantum effect it can produce macroscopic fields. For example the magnetic field of a ferromagnet is due to the magnetic moments of its unpaired electrons.

If you leave aside magnetic fields originating from quantum effects then yes, the observation of a magnetic field implies a non-zero current i.e. there must be charged particles in motion. To see this look at the derivation of the Biot-Savart law from Maxwell's equations. The magnetic field is the curl of the vector potential, and this is related to the current by:

$$ \nabla^2{\bf A} = -\mu_0 {\bf J} $$

If ${\bf J}$ is zero then there is no magnetic field.

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John Rennie
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  • So, you are looking magnetic field as a motion of a charged particle from the perspective of quantum theory. You have also mentioned that this quantum effect can produce macroscopic fields. Then, according to you, there can't be any magnetic field in free space as there is no electron. Is it so? – Subhra Jul 12 '14 at 08:13