Question

A magnetic field is applied to a freely floating uniform iron sphere with radius R = 2.00 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 μx−→


What is the sphere’s resulting angular speed ω?

Answer 1

`\omega=4.2704*10^-^5 rad/s`

Step-by-step explanation:

Angular Momentum Formula For atoms=
`L_(atom)=0.12Nm_(s)h`

Where:

m_{s}h is the momentum for one atom (m_s is the spin quantum number)

N is the number of atoms=
`(N_(A)*m)/(M)`

Where:

N_A is Avogadro Number

m is the mass of sphere

M is the molar mass of iron

Angular Momentum Formula For atoms will be=
`L_(atom)=0.12(N_(A)m)/(M) m_(s)h`

Angular Momentum of Sphere=
`L_(sphere)=I\omega`

where:

`I=(2mR^(2))/(5)`

So,Angular Momentum of Sphere=
`L_(sphere)=(2mR^(2))/(5)\omega`

Angular Momentum of sphere=Angular Momentum of atoms

`L_(sphere)`=
`L_(atom)`

`(2mR^(2))/(5)\omega`=
`0.12(N_(A)m)/(M) m_(s)h`

For iron,
`m_s =(1)/(2)`. So above equation will become:

`\omega=(0.12*5*N_(A)h)/(4*M*R^(2) )`

Where R=2mm, M=0.0558Kg/mol (Molar Mass of iron),h=Planck's Constant/2π

`\omega=(0.12*5*(6.022*10^(23))(6.63*10^(-34)/2*\pi))/(4*0.0558*(2*10^(-3))^(2))`

`\omega=4.2704*10^-^5 rad/s`