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Outline of proof that a magnetic field calculated by the Biot-Savart law will always satisfy Gauss's law for magnetism and Ampere's law . We start with the Biot-Savart law: Plugging in the well-known relation and using the product rule for curls, as well as the fact that J does not depend on the unprimed coordinates, this equation can be rewritten as Since the divergence of a curl is always zero, this establishes Gauss's law for magnetism . Next, taking the curl of both sides, using the formula for the curl of a curl (see the article Curl (mathematics) ), and again using the fact that J does not depend on the unprimed coordinates, we eventually get the result Finally, plugging in the relations (where δ is the Dirac delta function ), using the fact that the divergence of J is zero (due to the assumption of magnetostatics ), and performing an integration by parts, the result turns out to be i.e. Ampere's law . thus the ampere's circuital law is prooved


Outline of proof that a magnetic field calculated by the Biot-Savart law will always satisfy Gauss's law for magnetism and Ampere's law. We start with the Biot-Savart law:

\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int d^3r' \mathbf{J}(\mathbf{r}')\times \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}

Plugging in the well-known relation

\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} = -\nabla\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)

and using the product rule for curls, as well as the fact that J does not depend on the unprimed coordinates, this equation can be rewritten as

\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \nabla\times\int d^3r' \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}

Since the divergence of a curl is always zero, this establishes Gauss's law for magnetism. Next, taking the curl of both sides, using the formula for the curl of a curl (see the article Curl (mathematics)), and again using the fact that J does not depend on the unprimed coordinates, we eventually get the result

\nabla\times\mathbf{B} = \frac{\mu_0}{4\pi}\nabla\int d^3r' \mathbf{J}(\mathbf{r}')\cdot\nabla\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) - \frac{\mu_0}{4\pi}\int d^3r' \mathbf{J}(\mathbf{r}')\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)

Finally, plugging in the relations

\nabla\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) = -\nabla' \left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right),
\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) = -4\pi \delta(\mathbf{r}-\mathbf{r}')

(where δ is the Dirac delta function), using the fact that the divergence of J is zero (due to the assumption of magnetostatics), and performing an integration by parts, the result turns out to be

\nabla\times \mathbf{B} = \mu_0 \mathbf{J}

i.e. Ampere's law.

 

thus the ampere's circuital law is prooved


Grade:10

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