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How can to derive $F = m H$?

I know:

$$ F = \frac{\mu m_1 m_2}{4\pi r^2}$$

$$\mu = \frac {B}{H} $$

$$ H = \frac{\mu m}{4\pi r^2} $$

where '$F$' is Force, '$H$' is Magnetic Intensity, '$\mu$' is permeability, '$m$' is pole strength, and '$H$' is Magnetic Field Intensity.

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  • $\begingroup$ Are you sure it isn't $F=mH$? That would be consistent with the other things you know. $\endgroup$ May 20, 2020 at 6:21

2 Answers 2

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There is probably a typo in your expression: $F=mB$. It should be $F=mH$

Given $$F=\frac{\mu m_1m_2}{4\pi r^2}\tag1$$ $$\mu=\frac{B}{H}\tag 2$$ $$H=\frac{\mu m}{4\pi r^2} \tag3$$ Now, diving (1) by (3) (eq(2) is superfluous i.e. not required), $$\frac{F}{H}=\frac{\mu m_1m_2}{4\pi r^2}\cdot \frac{4\pi r^2}{\mu m}$$

$$ \frac FH=\frac{m_1m_2}{m}$$ Assuming magnetic dipole moments $m_1=m_2=m$, we get $$\frac FH=\frac{m\cdot m}{m}=m$$ $$\therefore F=mH$$

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As someone else pointed out, i think you mean F=mH.

The derivation is as followed

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  • $\begingroup$ Thanks a lot! It seems it was a typo in my book. 😊 $\endgroup$
    – Shub
    May 20, 2020 at 9:40
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    $\begingroup$ Scans of text and math are not considered acceptable on this site, either for questions or answers. They aren’t searchable, and they aren’t accessible by visually-impaired members. Please replace your scan with text and MathJax. $\endgroup$
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    May 20, 2020 at 17:12

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