Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Assume that Each Iron Atom Has a Permanent Magnetic Moment Equal to 2 Bohr Magnetons (1 Bohr Magneton Equals 9.27 × 10−24 A M2). the Density of Atoms in Iron is 8.52 × 1028 Atoms M−3. - Physics

Sum

Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magneton equals 9.27 × 10−24 A m2). The density of atoms in iron is 8.52 × 1028 atoms m−3. (a) Find the maximum magnetisation I in a long cylinder of iron (b) Find the maximum magnetic field B on the axis inside the cylinder.

#### Solution

Given:-

No of atoms per unit volume, f = 8.52 × 1028 atoms/m3

Magnetisation per atom, M = 2 × 9.27 × 10−24 A-m2

(a) Intensity of magnetisation, $I = \frac{M}{V}$

⇒ I = 2 × 9.27 × 10−24 × 8.52 × 1028

⇒ I = 1.58 × 106 A/m

(b) For maximum magnetisation ,the magnetising field will be equal to the intensity of magnetisation.

So, I = H

Magnetic field (B) will be,

B = 4π × 10−7 × 1.58 × 106

⇒ B ≈ 19.8 × 10−1 = 2.0 T

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 15 Magnetic Properties of Matter
Q 8 | Page 287