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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.