If magnetic field is conservative, then why not the magnetic force?

My professor thinks it is non conservative but he couldn't explain to me why?


This is because of the definition of a conservative force (and links therein):

If a force acting on an object is a function of position only, it is said to be a conservative force, and it can be represented by a potential energy function which for a one-dimensional case satisfies the derivative condition

conservative force

Lets look at the magnetic field, can it be described by a scalar potential?

There is no general scalar potential for magnetic field B but it can be expressed as the curl of a vector function

${\vec{B} = \vec{\nabla} \times \vec{A}}$

So it does not fall within the definition of conservative forces.

Another view:

A force field $F$, defined everywhere in space (or within a simply-connected volume of space), is called a conservative force or conservative vector field if it meets any of these three equivalent conditions:

  1. The curl of $F$ is zero: $$\nabla \times \vec{F} = 0.$$

  2. There is zero net work ($W$) done by the force when moving a particle through a trajectory that starts and ends in the same place: $$W \equiv \oint_C \vec{F} \cdot \mathrm{d}\vec r = 0.$$

  3. The force can be written as the negative gradient of a potential, $\Phi$: $$\vec{F} = -\nabla \Phi.$$

[Proof of equivalence omitted.]

The term conservative force comes from the fact that when a conservative force exists, it conserves mechanical energy. The most familiar conservative forces are gravity, the electric force (in a time-independent magnetic field, see Faraday's law), and spring force.

Many forces (particularly those that depend on velocity) are not force fields. In these cases, the above three conditions are not mathematically equivalent. For example, the magnetic force satisfies condition 2 (since the work done by a magnetic field on a charged particle is always zero), but does not satisfy condition 3, and condition 1 is not even defined (the force is not a vector field, so one cannot evaluate its curl). Accordingly, some authors classify the magnetic force as conservative,[3] while others do not.[4] The magnetic force is an unusual case; most velocity-dependent forces, such as friction, do not satisfy any of the three conditions, and therefore are unambiguously nonconservative.

So it is not so clear, as with conservation of energy and momentum :).

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  • $\begingroup$ +1 also; however, I can't quite see what you're referring to when you say conditions 2, 1 and 3. $\endgroup$ – gj255 Jun 9 '14 at 10:03
  • $\begingroup$ @gj255 it is from the wiki link before. I did not copy everything $\endgroup$ – anna v Jun 9 '14 at 11:03
  • $\begingroup$ I don't agree with the wiki on this. You can evaluate the curl of the magnetic force and I have done so in my answer. This shows that only stationary magnetic force fields are conservative, as expected since B-fields do no work. Conditions1 and 2 are satisfied and a magnetostatic potential can be written to satisfy 3. $\endgroup$ – Rob Jeffries Sep 25 '16 at 16:19

This is a strange one. The magnetic field is NOT conservative in the presence of currents or time-varying electric fields.

A conservative field should have a closed line integral (or curl) of zero. Maxwell's fourth equation (Ampere's law) can be written $$ \nabla \times {\bf B} = \mu_0 {\bf J} + \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t}\, , $$ so you can see this will equal zero only in certain cases.

Magnetic force is also only conservative in special cases. The force due to an electromagnetic field is written $$ {\bf F} = q{\bf E} + q{\bf v} \times {\bf B}$$

For this to be conservative then $\nabla \times {\bf F} = 0$ and $$ \nabla \times {\bf F} = q\nabla \times {\bf E} + q\nabla \times ({\bf v} \times {\bf B}) .$$ But from Faraday's law we know that $$ \nabla \times {\bf E} = - \frac{\partial {\bf B}}{\partial t}\, , $$ so, $$ \nabla \times {\bf F} = - q\frac{\partial {\bf B}}{\partial t} + q{\bf v}(\nabla \cdot {\bf B}) - q{\bf B}(\nabla \cdot {\bf v}) + ({\bf B}\cdot \nabla){\bf v} - ({\bf v}\cdot \nabla){\bf B}\, .$$ From the solenoidal law $\nabla \cdot {\bf B}=0$ always, and $\nabla \cdot {\bf v} = \partial/\partial t(\nabla \cdot {\bf r})=0$. Furthermore, $({\bf B}\cdot \nabla){\bf v} = ({\bf B}\cdot \frac{\partial}{\partial t} \nabla){\bf r} = 0$, so $$ \nabla \times {\bf F} = - q\left[\frac{\partial {\bf B}}{\partial t} + \frac{\partial {\bf B}}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial {\bf B}}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial {\bf B}}{\partial z} \frac{\partial z}{\partial t}\right] $$ $$\nabla \times {\bf F} = - q\frac{d {\bf B}}{d t}$$ and the force is only conservative in the case of stationary magnetic (and hence electric) fields.

Edit: Note that work is done by time-varying B-fields because of the inevitable accompanying E-field. So that may be a potential point of ambiguity.

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  • $\begingroup$ With your evaluation of curl of F, can you prove that $\oint \vec{F}\cdot d\vec{r}=0$? $\endgroup$ – mithusengupta123 Nov 25 '18 at 17:18
  • $\begingroup$ @mithusengupta123 But that equation isn't true if curl $F \neq 0$. $\endgroup$ – Rob Jeffries Nov 25 '18 at 18:07
  • $\begingroup$ But we know that the closed loop integral is zero indeed for magnetic Lorentz force. On the other hand, you showed that curl is nonzero. Are these two facts compatible? $\endgroup$ – mithusengupta123 Nov 25 '18 at 18:28
  • $\begingroup$ @mithusengupta123 The closed line integral of the Lorentz force is not zero in the presence of time-varying magnetic fields. That is the entire point of my answer. You cannot treat the electric and magnetic fields separately if the magnetic field is time-varying. $\endgroup$ – Rob Jeffries Nov 25 '18 at 18:30
  • $\begingroup$ Okay. The last line. I see it now. So the Lorentz force $\vec{F}$ in case of time-independent $\vec{B}$ field, in your opinion, conservative? It satisfies both 1 and 2 of anna v's answer. What about the 3? Can you define a scalar potential in that case for which $\vec{F}$ can be derived by taking the gradient? $\endgroup$ – mithusengupta123 Nov 25 '18 at 18:34

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