# Biot-Savart law

1. Aug 4, 2012

### erece

In the expression of Biot-Savart law
B = (µo/4π) ∫ (I dl x r^)/r2
why dl does not depend on the coordinate systems ?
in books they are using del X dl = 0

2. Aug 4, 2012

### Muphrid

It's natural $d\ell$ doesn't depend on the coordinate system; it's a vector.

Still, $\nabla \times d\ell$ doesn't make a whole lot of sense. Could you give some more context for this question?

3. Aug 5, 2012

### erece

i am attaching a file in which the derivation of biot-savart law is given. Now after equation 6-29 they used del X dl' = 0. i want to know the reason

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4. Aug 5, 2012

### Muphrid

It says the reason right in the screenshot:

Here, $d\ell$ and $d\ell'$ are two different things. The first depends on "unprimed" coordinates (i.e. $x, y, z$) and the second depends on "primed" coordinates $x', y', z'$). The operator $\nabla$ differentiates only with respect to the unprimed coordinates. Hence, $\nabla \times d\ell'$ is zero because $x', y', z'$ are not functions of $x,y, z$; they can't be differentiated with respect to the unprimed coordinates.

This use of primed and unprimed variables is pretty common in proofs of integral theorems, particularly when you have a function on the left that is, say, $X(r)$, generally the variable of integration on the right is $r'$. For example,

$$E(r) = \int \frac{\rho(r')/\epsilon_0}{4\pi |r - r'|^2} \; dV'$$

Here, $r'$ is the dummy variable of integration, and $dV'$ is the associated volume element for that variable.

5. Aug 5, 2012

### erece

now i got it. thanks a lot