# Does a displacement current have to be included in the Biot Savart Law?

Consider two spherical balls given charges of equal magnitude and opposite signs. Now, we connect them with a conducting straight finite wire. A current will flow in the wire till the charges on the spheres become zero. To find the magnetic field at a point lying on the perpendicular bisector of the wire, we may use Ampere's Law along with Maxwell's correction, since the electric field in changing with time in space.

How should I use Biot Savart Law to obtain the same expression for field? Should a displacement current be considered? Doesn't Biot Savart Law only give the field due to current flowing in conducting wires?

According to Griffiths and Heald, one can obtain a Biot-Savart-like formula $${\bf B} = \frac{\mu_0}{4 \pi} \int \frac{( {\bf J}+\epsilon_o \frac{\partial {\bf E}}{\partial t})\times \hat{{\bf r}}}{r^2} d \tau,$$ where the integrand is evaluated at the simultaneous (non-retarded) time. However, they point out that this is not useful for practical computation since it is self referential, namely to calcualte $${\bf B}$$ everywhere you need to know $${\bf E}$$ everywhere, and to know $${\bf E}$$ everywhere you need to know $${\bf B}$$ everywhere (because of Faraday induction, $${\bf \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}$$).

Heras commented that the more useful generalisation to the Biot Savart Law for time dependent sources was given by Jefimenko (Electricity and Magnetism, 2nd Ed., Electrect Scientific, Star City, 1989 p 516) and by Jackson (Classical Electrodynamics, 2nd Ed., Wiley, New York, 1975, p 225) and is given as $${\bf B} = \frac{\mu_0}{4 \pi} \int d^3x' \left( \frac{[{\bf J} \times {\hat {\bf R}}]}{R^2} + \frac{[\partial {\bf J}/\partial t] \times {\hat {\bf R}}}{Rc} \right),$$ where the square brackets indicate that the enclosed quantity is to be evaluated at the retarded time $$t'=t−R/c$$.

• – jim
Jul 14 '16 at 12:21
• Something is wrong with links at the end of this answer, when I click edit there are some references there but they do not show up on the page. Aug 18 '19 at 9:42
• @JánLalinský Not sure what happened there, have put in links
– jim
Aug 18 '19 at 10:37

Bio-savart's law is indeed a relationship between the current of a wire and it's magnetic field, a more exact form of lenz's law. To examine whether there is a displacement current correction term for the relationship and current and magnetic field, we have to look at the nature of magnetism in materials.

$$\nabla \times \bar{B} = \mu_{0}j + \epsilon_{0}\frac{d\bar{E}}{dt}$$

Split current density into it's constituent parts:

$$j = j_{f} + j_{P} + j_{M}$$

So there are current densities due to the free charge, the polarisation and the magnetisation.

$$j_{M} = \nabla \times M$$

$$j_{P} = \frac{d\bar{P}}{dt}$$

Adding in the terms and rearranging we can see where the material term for the magnetic field arises:

$$\nabla \times \bar{B} = \mu_{0}(j_{f} + j_{P} + j_{M}) + \epsilon_{0}\frac{d\bar{E}}{dt}$$

$$\nabla \times \frac{\bar{B}}{\mu_{0}} = (j_{f} + \frac{d\bar{P}}{dt} + \nabla \times M) + \epsilon_{0}\frac{d\bar{E}}{dt}$$

$$\nabla \times \frac{\bar{B}}{\mu_{0}} - M = j_{f} + \epsilon_{0}\frac{d\bar{E}}{dt} + + \frac{d\bar{P}}{dt}$$

$$(\frac{\bar{B}}{\mu_{0}} - \bar{M}) + \bar{H}$$

$$\epsilon_{0}\bar{E} + \bar{P} = \bar{D}$$

$$\epsilon_{0}\frac{d\bar{E}}{dt} + \frac{d\bar{P}}{dt} = \frac{d\bar{D}}{dt}$$

Therefore:

$$\nabla \times \bar{H} = j_{0} + \frac{d\bar{D}}{dt}$$

So as you can see it depends on the rate of change of the displacement current in the wire, which in turn in the rate of change of the electric field, or E.M.F added to the rate of change of the polarisation, or the net charge difference per unit space.