If there is a thin wire with current $I$ flowing through it, could I write the current density at all points in space of a horizontal 2D slice of the wire as $I \cdot \delta^2(\vec r)$ ?


I'm a bit new to the idea of the Dirac delta function, but as I understand, it can be used to model point sources. The area under the curve of the function is always 1. I'm not sure how to write it in 2-Dimensions so i'll write it as $\delta^2(\vec{r})$ where $\vec r$ is a 2D point in space.

Relationship between current & current density:

$$\iint_A{J\cdot dA}=I$$

Volume under the Dirac delta function is 1:

$$\iint_A I\cdot\delta^2(\vec r)\cdot dA=I\iint_A\delta^2(\vec r)\cdot dA=I$$

Setting the two equal to each other:

$$\iint_A{J\cdot dA}=\iint_A I\cdot\delta^2(\vec r)\cdot dA$$

$$J=I\cdot\delta^2 (\vec r)$$

Is this logic correct? The equality seems to work for calculating the $B$-Field around a long thin wire using Ampere's Law but I'm not sure if the math behind it is correct or if I'm just getting the right result by misunderstanding the math.

  • 1
    $\begingroup$ Look alright. But be aware that there will be a Jacobian factor when transform the current density to cylindric coordinate. $\endgroup$
    – ytlu
    Apr 13, 2021 at 4:06
  • $\begingroup$ Not a duplicate, but very closely related: Dirac Delta Magnetic field. In particular, see the comment on my answer there. $\endgroup$
    – Philip
    Nov 19, 2021 at 11:08
  • $\begingroup$ Assuming that your current density is homogeneous (which is rarely the case), this can be done. $\endgroup$
    – Roger V.
    Dec 21, 2021 at 10:35

3 Answers 3


You are thinking in the right direction, although it is not quite clear what $\delta^2(\vec{r})$ exactly means.

The current density of a current $I$ through a thin wire flowing along the $z$-axis can be written as $$\vec{J}(\vec{r})=I\delta(x)\delta(y) \vec{e}_z$$ where $\vec{e}_z$ is the unit vector in $z$-direction. You can check this by verifying $$\iint_A\vec{J}\cdot d\vec{A}=\begin{cases} I \quad\text{ , if the area $A$ includes the point $(x,y)=(0,0)$} \\ 0 \quad\text{ , if the area $A$ doesn't include the point $(x,y)=(0,0)$} \end{cases}$$


I am also trying to define a current density with delta function for a filament, wondering if it is correct. The definition is $$ \vec{j} = \frac{I}{\pi} \cdot \frac{\delta(\rho)}{\rho}\cdot \vec{e_z} $$ where $\rho$ is a radius in cylindrical coordinates. Thus if we calculate the current, it is $$ current = \int \vec{j}d\vec{S} = \int 2\pi I \cdot \frac{\delta(\rho)}{\rho}\cdot \vec{e_z} \cdot d\vec{S} $$ where the surface element is equal to $$ d\vec{S}=\vec{e_z}\rho d\rho d\phi $$ thus the current is equal to $$ \int \frac{I}{\pi} \cdot \frac{\delta(\rho)}{\rho}\cdot \vec{e_z} \cdot \vec{e_z}\rho d\rho d\phi = \frac{I}{\pi} \int_{0}^{\infty} \delta(\rho) d\rho \int_{0}^{2\pi}d\phi = \frac{I}{\pi}\frac{1}{2}2\pi=I $$

So it gives a proper current after integration


How to determine if a given charge and current density function satisfy continuity equation?

The general current density function is more complex what you have done only works for a straight wire located at x,y=0,0


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