# Biot–Savart law for discrete and continues current distribution

For the magnetic field generated from a wire in which current flows, we have:

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

In general tho for a continues current density, or for the magnetic field which is generated by some arbitrary object in which current flows, we have:

$$\vec B(\vec r)=\frac{\mu_0}{4\pi}\int_V \frac{\vec j(\vec r')\times(\vec r - \vec r')}{|\vec r- \vec r'|^3}dV'$$

From both equations we have the following implication:

$$I\cdot d\vec r'=\vec j(\vec r')dV$$

I tried to derive this last equation starting from the left side, but I cannot. This is where I am stuck:

$$I\cdot d\vec r'=j(\vec r') \cdot F \cdot d\vec r'=\vec j(\vec r') \cdot F \cdot dr'$$

The term $$F \cdot dr'$$ is not equal to $$dV'$$ because while $$dr'$$ is infinitesimal length, $$F$$ is the total surface and not an infinitesimal surface. So how can I derive the right side of my 3rd equation?

Well Unit analysis leads to the fact that they have the same units which indicates a connection. For a FULL derivation you need to express an infinitely thin current in the form of a VOLUME current, When integrating this function, the integral of J Dv reduces TO Idl, Jdv on its own cannot be used to prove this as they are technically different things

Well for a straight wire atleast in the z direction at x=0,y=0

The volume current density function for this would be $$J(r')=I_{0}\delta(x)\delta(y) \hat k$$

When x,y doesnt equal zero, the function is 0, when it is 0 the function is a particular infinite such that its integral is finite,

meaning when I integrate this function about a volume

$$\iiint I_{0}\delta(x)\delta(y) \hat k d^3r$$

$$\iiint I_{0}\delta(x)\delta(y) \hat k dxdydz$$

$$\iiint I_{0}\delta(y) \hat k dydz$$

$$\iiint I_{0}\hat kdz$$

notice this integral is just in the form Idz, a similar treatment for any shape wire would lead to dr

If there were denominators in the bottom the dirac would replace them with the correct values for a infinitely thin wire

The current density function is a must more complicated function for any shape wire.