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Is the electric force conservative in the presence of a magnetic field?

Maxwell's Equations tell us that the curl of $E$ is zero unless there is a changing magnetic field. Conservative forces are by definition zero curl.

Does this mean that if there is a changing magnetic field, the electric force stops being conservative? If so, does electric potential $V$ lose meaning? How do we handle this non-conservative aspect of the field? How does it not violate conservation of energy?

SRobertJames
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3 Answers3

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So:

$$ \vec E = -\nabla\phi $$

is an electrostatic formula for which the electric field is conservative. In a dynamic situation, there is another term:

$$ \vec E = -\nabla\phi -\frac{d\vec A}{dt}$$

that is time dependent. Moreover:

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

so if the vector potential is changing with time, so is the magnetic field.

So $\phi$ still exists, and is computed from charge density on the past light-cone (defined by $t_r = t-|\vec r - \vec r'|/c$):

$$ \phi(\vec r, t) = \frac 1 {4\pi\epsilon_0} \int\frac{\rho(\vec r', t_r)}{|\vec r -\vec r'|}dV' $$

and similarly for the vector potential and current density:

$$ \vec J(\vec r, t) = \frac {\mu_0} {4\pi} \int\frac{\vec J(\vec r', t_r)}{|\vec r -\vec r'|}dV' $$

I can't explain why it conserves energy, since I don't understand why you think it doesn't.

JEB
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As you say, the curl of an electric field is non-zero when a time dependent magnetic field is present, so the electric field becomes non-conservative and this does mean energy is not conserved. However this does not indicate any breakdown in the structure of physics, but rather that the energy has to come from an external source, which is generally the agent creating the time dependent field.

An obvious example of this is an electrical generator. We use a time dependent magnetic field to generate a non-conservative electric field in the conductor, and that field does work by generating a current in the conductor. That is the electrons flow in a loop out of the generator, through (for example) my computer, and back into the generator doing work along the way. The energy has to be supplied as mechanical energy by whatever is driving the generator.

John Rennie
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We have:

  • Electromagnetic force: $\bf F = q\bf E + q\bf v \times \bf B$
  • Faraday law: $\nabla \times E = -\frac {\partial \bf B}{\partial t}$
  • Curl of vector cross product: $\nabla \times (\bf f \times \bf g) = (\bf g . \nabla)\bf f - \bf g (\nabla . \bf f ) - (\bf f . \nabla)\bf g + \bf f (\nabla . \bf g)$
  • Solenoidal law: $\nabla ⋅ \bf B = 0$
  • By definition of velocity: $\nabla . \bf v = \frac {\partial}{\partial t}(\nabla . \bf r) = 0$
  • By definition of velocity: $(\bf B . \nabla)\bf v = (\bf B . \frac{\partial}{\partial t}\nabla)\bf r = 0$

Therefore, curl of electromagnetic force:

$\nabla \times \bf F$

$= q(\nabla \times \bf E + \nabla \times(\bf v \times \bf B))$

$= -q(\frac {\partial \bf B}{\partial t} + (\bf B . \nabla)\bf v - \bf B (\nabla . \bf v ) - (\bf v . \nabla)\bf B + \bf v (\nabla . \bf B))$

$= -q(\frac {\partial \bf B}{\partial t} + \frac{\partial B}{\partial x}\frac {\partial x}{\partial t} + \frac{\partial B}{\partial y}\frac {\partial y}{\partial t} + \frac{\partial B}{\partial z}\frac {\partial z}{\partial t})$

$= -q\frac {d \bf B}{d t}$

Thus, curl of electric force is zero in a steady (time-independent) magnetic field, but it is not zero in a changing (time-dependent) magnetic field.

A force is called conservative if work done by the force in moving between two points does not depend on the path, that is, if curl of the force is zero.

Therefore, electric force is conservative in a steady (time-independent) magnetic field, but it is not conservative in a changing (time-dependent) magnetic field.

Lalit Patel
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