In the course of "Classical Electrodynamics", our professor stated Biot-Savart's Law as follows: $$\vec {dB}=\frac{\mu_0}{4\pi}\cdot \frac{i\vec {dl} \times \vec r}{r^3}$$

Then he proceeded to derive the vector form of Biot-Savart's Law in the following way:

\begin{align} \vec {B}&=\frac{\mu_0}{4\pi}\cdot \int_C \frac{i\vec {dl} \times \vec r}{r^3}\\ &=\frac{\mu_0}{4\pi}\cdot \int_C \frac{(\iint \vec j\cdot \vec{ds})\vec {dl} \times \vec r}{r^3}\\ &=\frac{\mu_0}{4\pi}\cdot \iiint_V \frac{\vec j \times \vec r}{r^3} dV \end{align}

Is the derivation correct? The result is correct since in my book, the result is like this only. But I doubt whether this method or derivation is correct or not.

The reason for my thinking so is mentioned below:

We know that for any three vectors $\vec A,\vec B,\vec C$, $$(\vec A\cdot \vec B)\vec C \not = \vec A(\vec B\cdot \vec C)$$

But then in the derivation, we see that $$(\vec j\cdot \vec{ds})\vec {dl}=\vec j(\vec{ds}\cdot \vec {dl})=\vec j dV$$

But this is not true as per the above argument.

Am I correct? Or is the derivation correct?

Or is it so that the derivation is intuitive in that case and not absolutely mathematically rigorous?



1 Answer 1


You have to think about what $\vec j,\mathrm{d}\vec l$ and $\mathrm{d}\vec s$ actually are: $\mathrm{d}\vec l$ points along the flow of the current. So does $\vec j$. So $\mathrm{d}\vec l$ and $\vec j$ are parallel, and indeed $$ (\vec j\cdot\mathrm{d}\vec s) \mathrm{d}\vec l = (\mathrm{d}\vec l\cdot\mathrm{d}\vec s)\vec j$$ holds in that case.


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