These Q's are probably simple for those of you who have/completed a course in Vector Calculus. But I'm only a higschool/secondaryschool student, so I haven't.(adsbygoogle = window.adsbygoogle || []).push({});

1. I'm not sure how the infinitisemal magnetic flux density [itex]d\textbf{B}[/itex] from a wire element [itex]d\textbf{l}[/itex] with a current [itex]I[/itex], which is

[tex]d\textbf{B}=\frac{\mu_0I}{4\pi}\frac{d\textbf{l}\times\textbf{x}}{|\textbf{x}|^3}[/tex]

Can be written as

[tex]\textbf{B}(\textbf{x})=\frac{\mu_0}{4\pi}\int\textbf{J}(\textbf{x}')\times\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3}\,d^3x'[/tex]

Is it by using the definition of the current density,

[tex]I=\iint_A \textbf{J}\bullet\textbf{n}\,dA[/tex]

and insert this in the equation? but I can't figure it out, hope someone can help me on this one.

2. I know that

[tex]\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3} =

-\nabla\left(\frac{1}{|\textbf{x}-\textbf{x}'|}\right)[/tex]

It is then stated that by the use of this, you get

[tex]\textbf{B}(\textbf{x})=\frac{\mu_0}{4\pi}\,\nabla\times\int\frac{\textbf{J}(\textbf{x}')}{|\textbf{x}-\textbf{x}'|}\,d^3x'[/tex]

Is this because

[tex]-\nabla\left(\frac{1}{|\textbf{x}-\textbf{x}'|}\right)\times\textbf{J}(\textbf{x}')=\frac{-1}{|\textbf{x}-\textbf{x}'|}\left(\nabla\times\textbf{J}(\textbf{x}')\right)[/tex]

But what happens to the minus sign? Is it because that be take the gradient with respect to [itex]\textbf{x}[/itex] which is negative, and therefore when we take the curl of the current density we have to take it with respect to [itex]\textbf{x}'[/itex] which is negative, and therefore the gradient would be positive?

The book I'm using is

John David Jackson, Classical Electrodynamics, 3. ed

This subject can be found under 5.3, page 178.

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# Infinitisemal magnetic flux density

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