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A donut-shaped permanent magnet (magnetization parallel to the axis) can slide frictionlessly on a vertical rod. Treat the magnets as dipoles with mass $m_d$ and dipole moment $M$ . When we put two back to back magnets on the rod the upper one will float. At what height $z$ does it float?
${\left[ {\frac{{2{\mu _0}{M^2}}}{{3\pi {m_d}g}}} \right]^{1/4}}$
${\left[ {\frac{{6{\mu _0}{M^2}}}{{\pi {m_d}g}}} \right]^{1/4}}$
${\left[ {\frac{{3{\mu _0}{M^2}}}{{2\pi {m_d}g}}} \right]^{1/4}}$
${\left[ {\frac{{{\mu _0}{M^2}}}{{6\pi {m_d}g}}} \right]^{1/4}}$
Solution
Force between two magnetic dipoles parallel to each. $=\vec{\mathrm{F}}_{\mathrm{ab}}=\frac{3 \mu_{0}}{2 \pi r^{4}}\left(\vec{\mathrm{m}}_{\mathrm{e}} \overrightarrow{\mathrm{m}}_{\mathrm{b}}\right) \hat{\mathrm{r}}$
$|\overrightarrow{\mathrm{F}}|=\mathrm{mg}$
$\therefore \frac{3 \mu_{0}}{2 \pi r^{4}}\left(\mathrm{m}_{\mathrm{b}} \mathrm{m}_{\mathrm{a}}\right)=\mathrm{mg}$
$r=\left(\frac{3 \mu_{0} m^{2}}{2 \pi m g}\right)^{1 / 4}$
