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$$ d\vec{B}~=~\frac{\mu_0}{4\pi}\frac{I d\vec{\ell}\times \vec{r}}{r^3}. $$

If we change the $ d\vec{\ell}\times \vec{r} $ with the $ d\vec{\ell}\times$r . sin(Ө) = $ d\vec{\ell}\times$R in order to find the magnitude, then it means magnetic field falls off as the inverse cube of the distance not the inverse square, even the college physics books write it as an inverse square law.

Can we write the Biot-Savart formula with R, like this:

$$ d\vec{B}~=~\frac{\mu_0}{4\pi}\frac{I d\vec{\ell}.R}{r^3}. $$

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  • $\begingroup$ only if $\sin(\theta)=1$ $\endgroup$
    – Sigma
    Mar 18, 2015 at 12:22
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    $\begingroup$ even law of Gravitation can be written as $\vec{F}=\frac{Gm_1.m_2}{r^3}\vec{r}$ which is essentially $\vec{F}=\frac{Gm_1.m_2}{r^2}\hat{r}$ $\endgroup$
    – Sigma
    Mar 18, 2015 at 12:26
  • $\begingroup$ But it adds sin() to the equation unlike gravity. So the magnetic field falls off not just r^2 but (r^2 + sin). So it is not like gravitational field.Right? $\endgroup$
    – user50322
    Mar 18, 2015 at 12:38
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    $\begingroup$ $\sin$ is multiplied , I don't think its added $\endgroup$
    – Sigma
    Mar 18, 2015 at 13:00

1 Answer 1

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The distance on top, $R$ or $r$ effectively cancels the $r^3$ on the bottom to give overall a $r^{-2}$ factor, i.e. inverse square. Often you'll find expressions like this in EM, with things like: $$ {\vec r \over r} = \hat{r} $$ so that $$ {\vec r \over r^3} = {\hat{r} \over r^2} $$

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  • $\begingroup$ But while canceling, it adds sin() to the equation. So the magnetic field falls off not just r^2 but (r^2 + sin). Right? So it is not like gravitational field.Right? $\endgroup$
    – user50322
    Mar 18, 2015 at 12:37
  • $\begingroup$ the sin comes from the magnitude of the cross product, a way of expressing the relative directions of $r$ and $dl$ $\endgroup$
    – danimal
    Mar 18, 2015 at 13:15

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