Question

A magnetic field is applied to a freely floating uniform iron sphere with a radius R = 2.0 mm. The sphere initially had no net magnetic moment, but the field aligns 12% of the magnetic moments of the atoms (that is, 12% of the magnetic moments of the loosely bound electrons in the sphere, with one such electron per atom). The magnetic moment of those aligned electrons is the sphere’s intrinsic magnetic moment ${\overrightarrow{\mathbf{\mu }}}_{\mathbf{s}}$. What is the sphere’s resulting angular speed ?

Short answer

The resulting angular speed of the sphere is $4.27\times {10}^{-5}\mathrm{rad}/\mathrm{s}$.

Step-by-step solution

The given data:

a) The radius of the sphere, R = 2 mm

b) The magnetic field is applied to the uniform sphere.

c) Initially, it had no net magnetic moment.

d) But, the field aligns 12% of the magnetic moments of the atoms giving the sphere’s intrinsic magnetic moment as $\overrightarrow{{\mathrm{\mu }}_{\mathrm{s}}}$(with one such electron per atom).

Understanding the concept of magnetic dipole moment:

Using the concept of orbital spin angular momentum for N number of atoms, the required equation of the angular momentum in terms of mass. Now, the angular momentum due to rotation can be defined in terms of mass and angular velocity. As the net angular momentum in both cases is the same, equation both will be given the desired value of the angular speed.

Formulas:

The number of atoms present in the mass of the sphere,

$\mathrm{N}=\frac{{\mathrm{N}}_{\mathrm{A}}\mathrm{m}}{\mathrm{M}},{\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{\mathrm{mol}}^{-1}$ ….. (1)

The orbital angular spin momentum of an atom,

${\overrightarrow{\mathrm{L}}}_{\mathrm{atom}}={\mathrm{m}}_{\mathrm{s}}ħ$ ….. (2)

The angular momentum of the rotating sphere,

${\mathrm{L}}_{\mathrm{sphere}}=\left(\frac{2}{5}{\mathrm{mR}}^2\right)\mathrm{\omega }\left(\mathrm{Moment}\mathrm{of}\mathrm{interia}\mathrm{of}\mathrm{sphere},\mathrm{I}=\frac{2}{5}{\mathrm{mR}}^2\right)$ ….. (3)

Calculation of the angular speed of the sphere:

At first, using the molar mass of iron (M = 0.0558 kg/ mol ) in equation (1), the number of atoms present in the sphere is as follow.

$\mathrm{N}=\frac{\left(6.02\times {10}^{23}{\mathrm{mol}}^{-1}\right)\mathrm{m}}{\left(0.0558\mathrm{kg}/\mathrm{mol}\right)}\left(\because {\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{\mathrm{mol}}^{-1}\right)$

Now, the orbital angular momentum of N aligned atoms for the given condition that only 12% are aligned can be given using the above value in equation (2) as follows:

$$\begin{gathered} {\overrightarrow{\mathrm{L}}}_{\mathrm{atom}}=0.12{\mathrm{Nm}}_{\mathrm{s}}\mathrm{ħ} \\ {\overrightarrow{\mathrm{L}}}_{\mathrm{atom}}=\frac{\left(0.12\right)\left(6.02\times {10}^{23}{\mathrm{mol}}^{-1}\right)\mathrm{m}}{\left(0.0558\mathrm{kg}/\mathrm{mol}\right)}\left(\frac{\mathrm{ħ}}{2}\right)\left(\because {\mathrm{m}}_2=\frac{1}{2}\right) \end{gathered}$$

….. (4)

But, according to the question, the angular momentum of the aligned atoms is equation to the angular momentum of the sphere. Thus, comparing equations (4) and (3), the angular speed of the sphere as follows:

$$\begin{gathered} \frac{\left(0.12\right)\left(6.02\times {10}^{23}{\mathrm{mol}}^{-1}\right)\mathrm{m}}{0.0558\mathrm{kg}/\mathrm{mol}}\left(\frac{\mathrm{ħ}}{2}\right)=\left(\frac{2}{5}{\mathrm{mR}}^2\right)\mathrm{\omega } \\ \mathrm{\omega }=\frac{\left(0.12\right)\left(6.02\times {10}^{23}{\mathrm{mol}}^{-1}\right)\mathrm{m}}{0.0558\mathrm{kg}/\mathrm{mol}}\left(\frac{6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}/2\mathrm{\pi }}{2}\right)\left(\frac{5}{2{\left(0.002\mathrm{m}\right)}^2}\right) \\ =4.27\times {10}^{-5}\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the value of the angular speed is$4.27\times {10}^{-5}\mathrm{rad}/\mathrm{s}$