Relationship between Bit-Savart and Ampere laws

I've tried to relate Biot-Savart's Law to Ampere's and I've found a contradiction, which I guess is due to a naive use of Ampere's.

If ## \int \vec B · d\vec l = \mu_0 I_{enc} ## is applied to a circle of radius R around a current element ##Id\vec l## we have ## B·2\pi R = \mu_{0} I_{enc} ##, which gives ## B = \frac{\mu_0 I_{enc}}{2\pi R} ##, different from Biot-Savart. I'm guessins we can't use AMpere in this situation, but I can't put my finger on the exact reason.

Mende

Dale
Mentor
2020 Award
An isolated current element is impossible, it does not conserve charge. Instead of a current element you need to use a loop. Try an infinitely long straight wire with a return path at infinity.

carllacan
jtbell
Mentor
If B⃗ ·dl⃗ =μ0Ienc \int \vec B · d\vec l = \mu_0 I_{enc} is applied to a circle of radius R around a current element IdlId\vec l
One problem here is with the concept of Ienc which is often stated in first-year textbooks as "the current through the loop" (circle in this case). What this means precisely is: "the current that passes through any surface whose boundary is the loop." The problem is that with a finite current element, you can construct some surfaces that the current "pierces", and some that that the current does not "pierce" (i.e. the surface is curved in such a way as to avoid the current element entirely). So "current through the loop" is ambiguous for a finite current element. It depends on which surface you use. If you use a complete current loop instead of a finite element, you avoid the ambiguity.

As DaleSpam also noted, an isolated current element (or any finite-length current-carrying wire) does not conserve charge all by itself. You can make it conserve charge by attaching a charged object to each end of the wire. As the current flows, one charge becomes more negative and the other becomes more positive. As these charges change, the electric field that they produce also changes. This changing electric field is associated with a magnetic field (in addition to the magnetic field associated with the current), according to the term that Maxwell added to Ampere's Law in order to make his equations "complete." This term compensates for the different possible values of "current through the loop".
$$\oint {\vec B \cdot d \vec l} = \mu_0 \int {\vec J \cdot d \vec a} + \mu_0 \epsilon_0 \frac {d}{dt} \int {\vec E \cdot d \vec a} \\ \oint {\vec B \cdot d \vec l} = \mu_0 i_\textrm{enc} + \mu_0 \epsilon_0 \frac {d}{dt} \int {\vec E \cdot d \vec a}$$

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Hechima and Dale
Wow, that was enlightening. Many thanks you both!

vanhees71
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{J} + \frac{1}{c^2} \partial_t \vec{E}.$$
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{B}=\mu_0 \vec{J} + \frac{1}{c^2} \int_A \mathrm{d}^2 \vec{a} \cdot \partial_t \vec{E}.$$
$$\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{a} \cdot \vec{E}=\int_{A} \mathrm{d}^2 \vec{a} \cdot \left [\partial_t \vec{E}+\vec{v} (\vec{\nabla} \cdot \vec{E}) \right ] - \int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{E}).$$