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 m_d and dipole moment m.

(a) If you put two back-to-hack 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.) $\left[Answer:\left(a\right)3{\mu }_0{m}^2\left(b\right)0.8501\right]$

Short answer

(a) The height at which the upper magnet floats is ${\left(\frac{3{\mu }_0{m}^2}{2{\mathrm{\pi m}}_{\mathrm{d}}\mathrm{g}}\right)}^{\frac{1}{4}}$.

(b) The ratio for height is 0.85011.

Step-by-step solution

Write the given data from the question.

Dipole mass is m_d.

Dipole moment is m.

Determine the height at which upper floats.

The expression for the magnetic field due upper magnet is given by,

$\overrightarrow{{\mathbf{B}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{m}}_{\mathbf{1}}}{\mathbf{4}{\mathbf{\pi z}}^{\mathbf{2}}}\left(2cos\theta \hat{r}+sin\theta \hat{\theta }\right)$ …… (1)

Here, m₁ is the magnetic dipole moment, z is the distance where the magnetic is located.

Substitute 0 for $\theta$into equation (1).

localid="1657709952752" $$\begin{gathered} \overrightarrow{B_1}=\frac{{\mu }_0{m}_1}{4{\mathrm{\pi z}}^3}\left(2\cos 0\stackrel{⏜}{z}+\sin 0\hat{\theta }\right) \\ \overrightarrow{B_1}=\frac{{\mu }_0{m}_1}{4{\mathrm{\pi z}}^3}\left(2\hat{z}\right) \\ \overrightarrow{B_1}=\frac{{\mu }_0{m}_1}{2{\mathrm{\pi z}}^3}\hat{z} \end{gathered}$$

The expression for the magnetic field due lower magnet is given by,

$\overrightarrow{B_2}=m_2\overrightarrow{B_1}$ …… (2)

Here m₂ is the magnetic dipole moment.

Substitute$\frac{{\mu }_0{m}_1}{2{\mathrm{\pi z}}^3}\hat{z}$for$\overrightarrow{B_1}$into equation (2)

localid="1657710275896" $\overrightarrow{B_2}=m_2\frac{{\mu }_0{m}_1}{2{\mathrm{\pi z}}^3}\hat{z}$ …… (3)

The magnetic dipole moment of both the magnate is equal. Therefore,

m₁ = m₂

m₂ = m

Substitute m for m₁ and m₂ into equation (3).

$$\begin{gathered} \overrightarrow{B_2}=m\frac{{\mu }_0{m}_1}{2{\mathrm{\pi z}}^3}\hat{z} \\ \overrightarrow{B_2}=\frac{{\mu }_0{m}_1}{2{\mathrm{\pi z}}^3}\hat{z} \end{gathered}$$

The magnetic force on an infinitesimal loop is given by,

$\overrightarrow{F}=\nabla \left(\overrightarrow{B_2}\right)$

Substitute $\frac{{\mu }_0{m}^2}{2{\mathrm{\pi z}}^3}\hat{z}$for $\overrightarrow{B_2}$into above equation.

localid="1657710798901" $$\begin{gathered} \overrightarrow{F}=\frac{\partial }{\partial z}\left(\frac{{\mu }_0{m}^2}{2{\mathrm{\pi z}}^3}\hat{z}\right) \\ \overrightarrow{F}=\frac{3{\mu }_0{m}^2}{2{\mathrm{\pi z}}^3}\hat{z} \end{gathered}$$

The weight of the upper magnet is given by,

W = m_dg

Here m_d is the mass of the upper magnet and g is the acceleration due to gravity.

The upper magnet is balanced by the weight of magnet.

F = W

Substitute $\frac{3{\mu }_0{m}^2}{2{\mathrm{\pi z}}^4}$for $\overrightarrow{F}$ and m_dg for W into above equation.

$$\begin{gathered} \frac{3{\mu }_0{m}^2}{2{\mathrm{\pi z}}^4}=m_{d}g \\ {z}^4=\frac{3{\mu }_0{m}^2}{2\pi m_{d}g} \\ z={\left(\frac{3{\mu }_0{m}^2}{2\pi m_{d}g}\right)}^{\frac{1}{4}} \end{gathered}$$

Hence the height at which the upper magnet floats is ${\left(\frac{3{\mu }_0{m}^2}{2\pi m_{d}g}\right)}^{\frac{1}{4}}$.

Calculate the ratio of the two heights.

Now the third magnet is added. The lower magnet repels the middle magnate in upward direction and upper magnet repels in downward direction.

The force on the middle magnet due to lower magnet is given by,

$F_{ml}=\frac{3{\mu }_0{m}^2}{2{\mathrm{\pi x}}^4}$

Here m stand for middle and I stand for lower.

The force on the middle magnet due to upper magnet is given by

$F_{mu}=\frac{3{\mu }_0{m}^2}{2\pi y^4}$

Here m stand for middle and u stand for upper.

The weight of the middle magnet is given by,

$W_m=m_{d}g$

The net force on the middle magnet is given by,

$F_{ml}=F_{mu}+W_m$

Substitute $m_{d}g$for W_middle,localid="1657711842584" $\frac{3{\mu }_0{m}^2}{2\pi y^4}$ for F_upper and $\frac{3{\mu }_0{m}^2}{2{\mathrm{\pi x}}^4}$for F_lower into above equation.

localid="1657711909907" width="175">3μ0m22πx4=3μ0m22πy4+mdg

$\frac{3{\mu }_0{m}^2}{2{\mathrm{\pi x}}^4}-\frac{3{\mu }_0{m}^2}{2\pi y^4}-m_{d}g=0$ ……. (4)

The force on the upper magnet due to lower magnet is given by,

$F_{ul}=\frac{3{\mu }_0{m}^2}{2\pi {(x+y)}^4}$

The force on the upper magnet due to middle magnet is given by,

$F_{um}=\frac{3{\mu }_0{m}^2}{2\pi y^4}$

The weight of upper magnet is given by,

W_u= m_dg

The net force on upper magnet is given by,

$F_{um}=F_{ul}+W_u$

Substitute$\frac{3{\mu }_0{m}^2}{2\pi {(x+y)}^4}$ for F_ul,$\frac{3{\mu }_0{m}^2}{2\pi y^4}$ for F_um and m_dg for W_u into above equation.

role="math" localid="1657714396739" $\frac{3{\mu }_0{m}^2}{2\pi y^4}=\frac{3{\mu }_0{m}^2}{2\pi {(x+y)}^4}+m_{d}g$

$\frac{3{\mu }_0{m}^2}{2\pi y^4}-\frac{3{\mu }_0{m}^2}{2\pi {(x+y)}^4}-m_{d}g=0$ …… (5)

Subtract the equation (5) from equation (4).

$\begin{array}{rcl}\left(\frac{3{\mu }_0{m}^2}{2\pi x^4}-\frac{3{\mu }_0{m}^2}{2\pi y^4}-m_{d}g\right)-\left(\frac{3{\mu }_0{m}^2}{2\pi y^4}-\frac{3{\mu }_0{m}^2}{2\pi {(x+y)}^4}-m_{d}g\right)& =& 0\\ \frac{3{\mu }_0{m}^2}{2\pi x^4}-2\left(\frac{3{\mu }_0{m}^2}{2\pi y^4}\right)+\frac{3{\mu }_0{m}^2}{2\pi {(x+y)}^4}& =& 0\\ \frac{3{\mu }_0{m}^2}{2\pi }\left(\frac{1}{x^4}-\frac{2}{y^4}+\frac{1}{{(x+y)}^4}\right)& =& 0\\ \left(\frac{1}{x^4}-\frac{2}{y^4}+\frac{1}{{(x+y)}^4}\right)& =& 0\end{array}$

Solve further as

$\begin{array}{rcl}\frac{1}{{\left({\displaystyle \frac{x}{y}}\right)}^4}-\frac{2}{1}+\frac{1}{{\left({\displaystyle \frac{x}{y}}+1\right)}^4}& =& 0\\ \frac{1}{{\left({\displaystyle \frac{x}{y}}\right)}^4}+\frac{1}{{\left({\displaystyle \frac{x}{y}+1}\right)}^4}& =& 2\\ \frac{x}{y}& =& 0.85011\end{array}$

Hence the ratio for height is 0.85011.