# Confusion in understanding Biot-Savart law

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?

• 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! – Sebastian Riese Feb 20 '16 at 21:40
• 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. – garyp Feb 20 '16 at 21:42
• 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. – Bobak Hashemi Feb 21 '16 at 0:08

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 $I\mathrm d \vec l'=\hat J (J\mathrm da) \mathrm d l'=\vec J \mathrm d\tau'.$