A familiar toy consists of donut-shaped permanent magnets (magnetization parallel to the axis), which slide frictionlessly on a vertical rod (Fig. 6.31). Treat the magnets as dipoles, with mass md and dipole moment m.(a) If you put two back-to-back magnets on the rod, the upper one will “float”

Question

A familiar toy consists of donut-shaped permanent magnets (magnetization parallel to the axis), which slide frictionlessly on a vertical rod (Fig. 6.31). Treat the magnets as dipoles, with mass [latex]m_{d}[/latex] and dipole moment m.

(a) If you put two back-to-back magnets on the rod, the upper one will “float”—the magnetic force upward balancing the gravitational force downward. At what height (z) does it float?

(b) If you now add a third magnet (parallel to the bottom one), what is the ratio of the two heights? (Determine the actual number, to three significant digits.) [Answer: [latex] ext { (a) }\left[3 \mu_{0} m^{2} / 2 \pi m_{d} g ight]^{1 / 4} [/latex] ; (b) 0.8501]

Accepted Answer

(a) [latex]B _{1}=\frac{\mu_{0}}{4 \pi} \frac{2 m}{z^{3}} \hat{ z }( ext { Eq. } 5.88, ext { with } heta=0) . ext { So } m _{2} \cdot B _{1}=-\frac{\mu_{0}}{2 \pi} \frac{m^{2}}{z^{3}} . F = abla( m \cdot B )( ext { Eq. } 6.3) \Rightarrow F =\frac{\partial}{\partial z}\left[-\frac{\mu_{0}}{2 \pi} \frac{m^{2}}{z^{3}} ight] \hat{ z }=\frac{3 \mu_{0} m^{2}}{2 \pi z^{4}} \hat{ z }[/latex] . This is the magnetic force upward (on the upper magnet); it balances the gravitational force downward [latex]\left(-m_{d} g \hat{ z } ight)[/latex] :

[latex]B _{ ext {dip }}( r )= abla imes A =\frac{\mu_{0} m}{4 \pi r^{3}}(2 \cos heta \hat{ r }+\sin heta \hat{ heta })[/latex] (5.88)

[latex]\frac{3 \mu_{0} m^{2}}{2 \pi z^{4}}-m_{d} g=0 \Rightarrow z=\left[\frac{3 \mu_{0} m^{2}}{2 \pi m_{d} g} ight]^{1 / 4}[/latex] .

(b) The middle magnet is repelled upward by lower magnet and downward by upper magnet:

[latex]\frac{3 \mu_{0} m^{2}}{2 \pi x^{4}}-\frac{3 \mu_{0} m^{2}}{2 \pi y^{4}}-m_{d} g=0[/latex] .

The top magnet is repelled upward by middle magnet, and attracted downward by lower magnet:

[latex]\frac{3 \mu_{0} m^{2}}{2 \pi y^{4}}-\frac{3 \mu_{0} m^{2}}{2 \pi(x+y)^{4}}-m_{d} g=0[/latex] .

Subtracting: [latex]\frac{3 \mu_{0} m^{2}}{2 \pi}\left[\frac{1}{x^{4}}-\frac{1}{y^{4}}-\frac{1}{y^{4}}+\frac{1}{(x+y)^{4}} ight]-m_{d} g+m_{d} g=0, ext { or } \frac{1}{x^{4}}-\frac{2}{y^{4}}+\frac{1}{(x+y)^{4}}=0, ext { so: } 2=\frac{1}{(x / y)^{4}}+\frac{1}{(x / y+1)^{4}}[/latex] .

Let [latex]\alpha \equiv x / y ; ext { then } 2=\frac{1}{\alpha^{4}}+\frac{1}{(\alpha+1)^{4}}[/latex] . Mathematica gives the numerical solution [latex]\alpha=x / y=0.850115 \ldots[/latex]

Question 6.23: A familiar toy consists of donut-shaped permanent magnets (m...

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