The Biot-Savart law is

$${\bf B}( {\bf r} ) = \frac{\mu_0}{4 \pi} \int I \frac{{d \bf l'} \times (r-r')}{|r-r'|^3}$$

In Griffiths, for surface and volume currents the Biot-Savart law becomes

$${\bf B}( {\bf r} ) = \frac{\mu_0}{4 \pi} \int \frac{{\bf K}({\bf r'}) \times (r-r')}{|r-r'|^3} da'$$ $${\bf B}( {\bf r} ) = \frac{\mu_0}{4 \pi} \int \frac{{\bf J}({\bf r'}) \times (r-r')}{|r-r'|^3} d\tau'$$

It's very absurd. We already know that $I = \int {\bf J} \cdot d{\bf a}$. Also, Griffiths wrote that $${\bf K} \equiv \frac{d{\bf I}}{dl_\perp}, J \equiv \frac{d{\bf I}}{da_\perp}$$

I think we should fix the Biot-Savart form in this way. $${\bf B}( {\bf r} ) = \frac{\mu_0}{4 \pi} \int \frac{{\bf K}({\bf r'}) \times (r-r')}{|r-r'|^3} \cdot d{\bf l'}$$ $${\bf B}( {\bf r} ) = \frac{\mu_0}{4 \pi} \int \frac{{\bf J}({\bf r'}) \times (r-r')}{|r-r'|^3} \cdot d{\bf a'}$$

However the text doesn't explain like this. What is wrong here?

  • $\begingroup$ The relation between current density and surface current density is simply $\vec J(\vec r) = \int_S d\sigma(r')\, \delta(\vec r - \vec r') \vec K(\vec r')$ (where $\sigma$ is the surface measure) and for a current in a thin wire you get $\vec J = I \int_\gamma d\vec r'\,\delta(\vec r - \vec r')$. Simply plugging this into the law for current densities gives the results from Griffiths. Note that the resulting integrations are over surface/volume elements! $\endgroup$ – Sebastian Riese Feb 20 '16 at 21:40
  • $\begingroup$ You don't explain why you object to Griffiths. But check the units. Both ${\bf{K}}da$ and ${\bf{J}}d\tau$ have to have units of A$\cdot$ m. and they do. $\endgroup$ – garyp Feb 20 '16 at 21:42
  • $\begingroup$ I'm not totally sure what your confusion is in particular, but physics books do tend to gloss over serious mathematical issues like extending those integrals into higher dimensions. One thing I will say about your proposed forms for $\vec{B}$ is that the dimensionality doesn't work out. $\vec{B}$ is a vector, yet your integrand is a scalar. Perhaps that will help you find a more pointed question to ask? One other thing to consider is that integral on the first line takes $I$ to be 1D, whereas the $I = \int \vec{J}\cdot d\vec{a}$ is defining the current through a x-sectional area. $\endgroup$ – Bobak Hashemi Feb 21 '16 at 0:08

firstly, correct has direction but only through an infinitesimal area or volume, in anything bigger, it is not a vector as it doesn't follow vector addition, and it's direction is same as direction of density not anything to do with the volume or surface we choose, also if we would have taken dot product, magnetic field would have become scalar which is obviously not valid


Clearly your formulas have the wrong units.

Consider $I\mathrm d \vec l'=\vec I \mathrm d l'=\hat J I \mathrm d l'.$

If $I=\vec J\cdot\mathrm d\vec a'=J\mathrm da'$ then we get $ I\mathrm d \vec l'=\hat J (J\mathrm da) \mathrm d l'=\vec J \mathrm d\tau'.$


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