Question

what is biot savart law ans mi fast​

Answer 1

Answer:In a magnetostatic situation, the magnetic field B as calculated from the Biot–Savart law will always satisfy Gauss's law for magnetism and Ampère's law:[10]

Outline of proof[10] (Click "show" on the right.)

Starting with the Biot–Savart law:

{\displaystyle \mathbf {B} (\mathbf {r} )={\frac {\mu _{0}}{4\pi }}\iiint _{V}d^{3}l\mathbf {J} (\mathbf {l} )\times {\frac {\mathbf {r} -\mathbf {l} }{|\mathbf {r} -\mathbf {l} |^{3}}}} {\mathbf B}({\mathbf r})={\frac {\mu _{0}}{4\pi }}\iiint _{V}d^{3}l{\mathbf J}({\mathbf l})\times {\frac {{\mathbf r}-{\mathbf l}}{|{\mathbf r}-{\mathbf l}|^{3}}}

Substituting the relation

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

and using the product rule for curls, as well as the fact that J does not depend on {\displaystyle \mathbf {r} } \mathbf {r} , this equation can be rewritten as[10]

{\displaystyle \mathbf {B} (\mathbf {r} )={\frac {\mu _{0}}{4\pi }}\nabla \times \iiint _{V}d^{3}l{\frac {\mathbf {J} (\mathbf {l} )}{|\mathbf {r} -\mathbf {l} |}}} {\mathbf B}({\mathbf r})={\frac {\mu _{0}}{4\pi }}\nabla \times \iiint _{V}d^{3}l{\frac {{\mathbf J}({\mathbf l})}{|{\mathbf r}-{\mathbf l}|}}

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, and again using the fact that J does not depend on {\displaystyle \mathbf {r} } \mathbf {r} , we eventually get the result[10]

{\displaystyle \nabla \times \mathbf {B} ={\frac {\mu _{0}}{4\pi }}\nabla \iiint _{V}d^{3}l\mathbf {J} (\mathbf {l} )\cdot \nabla \left({\frac {1}{|\mathbf {r} -\mathbf {l} |}}\right)-{\frac {\mu _{0}}{4\pi }}\iiint _{V}d^{3}l\mathbf {J} (\mathbf {l} )\nabla ^{2}\left({\frac {1}{|\mathbf {r} -\mathbf {l} |}}\right)} \nabla \times {\mathbf B}={\frac {\mu _{0}}{4\pi }}\nabla \iiint _{V}d^{3}l{\mathbf J}({\mathbf l})\cdot \nabla \left({\frac {1}{|{\mathbf r}-{\mathbf l}|}}\right)-{\frac {\mu _{0}}{4\pi }}\iiint _{V}d^{3}l{\mathbf J}({\mathbf l})\nabla ^{2}\left({\frac {1}{|{\mathbf r}-{\mathbf l}|}}\right)

Finally, plugging in the relations[10]

{\displaystyle \nabla \left({\frac {1}{|\mathbf {r} -\mathbf {l} |}}\right)=-\nabla _{l}\left({\frac {1}{|\mathbf {r} -\mathbf {l} |}}\right),} \nabla \left({\frac {1}{|{\mathbf r}-{\mathbf l}|}}\right)=-\nabla _{l}\left({\frac {1}{|{\mathbf r}-{\mathbf l}|}}\right),

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

(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[10]

{\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} } \nabla \times {\mathbf B}=\mu _{0}{\mathbf J}

i.e. Ampère's law. (Due to the assumption of magnetostatics, {\displaystyle \partial \mathbf {E} /\partial t=\mathbf {0} } {\displaystyle \partial \mathbf {E} /\partial t=\mathbf {0} }, so there is no extra displacement current term in Ampère's law.)

In a non-magnetostatic situation, the Biot–Savart law ceases to be true (it is superseded by Jefimenko's equations), while Gauss's law for magnetism and the Maxwell–Ampère law are still true.

Explanation:

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