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I am supposed to prove: $$B_y=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi} \frac{\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3yz}{r^2}\biggl)$$ and $$B_z=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi} \frac{a-y\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3z^2}{r^2}-1\biggl)$$

One of the steps to doing it is proving: $$(a^2+y^2+z^2-2ay\sin\phi)^{3/2}\approx r^{-3}\biggl(1-\frac{2ay\sin\phi}{r^2}\biggl)^{-3/2}\approx r^{-3}\biggl(1+\frac{3ay\sin\phi}{r^2}\biggl)$$ using the Binomial theorem (the specific instruction reads "for this step use the binomial theorem, expand up to second term"). How can I go about this step, taking into account the following? $$r=\sqrt{z^2+y^2}>>a$$

David1607
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1 Answers1

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First note that:

  • $r = \sqrt{z^2+y^2} \gg a \implies z^2+y^2 \gg a^2 \implies a^2+y^2+z^2 \approx y^2+z^2$

  • $r \gt y \;\land\; r \gg a \implies r^2 \gg ay \implies r^2 \gg ay \sin \phi$

Then:

$$ \require{cancel} \begin{align} (a^2+y^2+z^2-2ay\sin\phi)^{-3/2} &\approx (y^2+z^2-2ay\sin\phi)^{-3/2} \\ &= \left(r^2 \left(1-\frac{2ay\sin\phi}{r^2}\right)\right)^{-3/2} \\ &= \left(r^{2}\right)^{-3/2} \cdot \left(1- \left(-\frac{3}{\cancel{2}}\right) \cdot \frac{\cancel{2}ay\sin\phi}{r^2} + \dots\right) \\ &\approx r^{-3}\biggl(1+\frac{3ay\sin\phi}{r^2}\biggl) \end{align} $$

dxiv
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