3
$\begingroup$

Can a magnetic field be induced without an electric field? Because, as far as I know, a time varying electric field induces a magnetic field an vice versa. But in the case of conductors carrying currennt, it doesn't seem that electric field varies with time, then how is a magnetic field induced?

$\endgroup$
1
  • 1
    $\begingroup$ If you hold a permanent magnet over a peace of metal, will that induce a magnetic field? $\endgroup$ Commented Feb 26, 2019 at 6:04

2 Answers 2

6
$\begingroup$

One of Maxwell’s four equations for electromagnetism in a vacuum shows how magnetic fields are produced:

$$\nabla\times\mathbf{B}=\frac{1}{c}\left(4\pi\mathbf{J}+\frac{\partial\mathbf{E}}{\partial t}\right).$$

(I’ve written it in Gaussian units.)

From this equation you can see that there are two different sources for magnetic fields: the first is a current density, and the second is a changing electric field.

So to have a magnetic field you do not need to have a time-varying electric field. You can just have moving charge. But when a magnetic field is produced by moving charge, physicists don’t call it “induced”.

$\endgroup$
2
  • 1
    $\begingroup$ But aren't moving charges producing an electric field? And what about spin? Can a neutron for instance create a B field while no electric field? $\endgroup$ Commented Feb 26, 2019 at 10:24
  • 2
    $\begingroup$ Moving charges don’t necessarily produce an electric field. For example you can have a positive charge density moving to the right and a negative charge density moving to the left at the same speed. There would be no net charge density but there would be a net current density. Quantum spin is another source of megnetic field. It is not in Maxwell’s equations because they are classical, not quantum. $\endgroup$
    – G. Smith
    Commented Feb 26, 2019 at 17:03
1
$\begingroup$

From Griffiths, Electrodynamics, Jefimenko’s equations are given as $${\bf E}({\bf r},t) = \frac{1}{4 \pi \epsilon_0} \int [ \frac{\rho ({\bf r}',t_r)}{{\mathfrak r}^2} {\bf \hat{\mathfrak r}} + \frac{\dot{\rho} ({\bf r}',t_r)}{c {\mathfrak r}} {\bf \hat{\mathfrak r}} - \frac{{\bf {\dot J}} ({\bf r}',t_r)}{c^2 {\mathfrak r}}] d \tau',$$ $${\bf B}({\bf r},t) = \frac{\mu_0}{4 \pi} \int [\frac{{\bf {J}} ({\bf r}',t_r)}{{\mathfrak r}^2} + \frac{{\bf {\dot J}} ({\bf r}',t_r)}{c {\mathfrak r}} ] \times {\bf \hat{\mathfrak r}} d \tau'.$$

These equations show that to create a magnetic field you require either a steady current or/as well a changing current. If the current density is steady (so that ${\bf {\dot J}} \equiv 0$) then you can see that you can arrange for no electric field by having the charge density $\rho$ vanish everywhere. Another way to create a magnetic field is to have a time varying current density, which necessarily creates an electric field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.