Question

Consider the situation when almost all of the magnetic moments of a sample of a particular ferromagnetic metal are aligned. In this case, the magnetic field can be calculated as the permeability constant μ0 multiplied by the magnetic moment per unit volume. In a sample of iron, for example, where the number density of atoms is approximately 8.50 ✕ 1028 atoms/m3, the magnetic field can reach 1.99 T. If each electron contributes a magnetic moment of 9.27 ✕ 10−24 A · m2 (1 Bohr magneton), how many electrons per atom contribute to the saturated field of iron?

Answer 1

Answer:

2.00976

Explanation:

B = Magnetic field = 1.99 T

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

$\mu_b$ = Magnetic moment of each electron = $9.27\times 10^{-24}\ Am^2$

n = Number density of atoms = $8.5\times 10^{28}\ atoms/m^3$

N = Number of electrons per atom

Magnetic field is given by

$B=N\mu_0\mu_bn\\\Rightarrow N=\dfrac{B}{\mu_0\mu_bn}\\\Rightarrow N=\dfrac{1.99}{4\pi\times 10^{-7}\times 9.27\times 10^{-24}\times 8.5\times 10^{28}}\\\Rightarrow N=2.00976$

The number of electrons per atom contribute to the saturated field of iron is 2.00976