Suppose each iron atom possesses a permanent magnetic moment of 2 Bohr magnetons (where 1 Bohr magneton equals 9.27 imes 10⁻²⁴ ext A m ^2 ). The density of atoms in iron is given as 8.52 imes 10²⁸ ext atoms m ⁻³ . Determine the maximum magnetic field B on the axis inside a long cylinder of iron.

Step-by-step solution

The maximum magnetization M is achieved when all atomic magnetic moments are aligned parallel to each other. M = n m Substituting the given values: n = 8.52 10²⁸ m ⁻³ m = 2 9.27 10⁻²⁴ A m ^2 = 1.854 10⁻²³ A m ^2 M = (8.52 10²⁸) (1.854 10⁻²³) 1.58 10^6 A m ⁻¹ For a long cylinder, the demagnetizing factor is zero, so the maximum magnetic field on the axis is given by B = ₀ M . B = 4 10⁻⁷ 1.58 10^6 B = 4 3.1416 0.158 1.98 T Rounding to the nearest given option, we get B 2.0 T .