Anna university solved problems

ANNA UNIVERSITY SOLVED PROBLEMS

 

Problem 2.12

The magnetic field strength of copper is10⁶ ampere/metre. If the magnetic susceptibility of copper is -0.8 × 10^-5, calculate the magnetic flux density and magnentisation in copper. (A.U. May 2014)

 

Given data

Magnetic field strength H = 10⁶ Am^-1

Magnetic susceptibility of copper χ = -0.8 × 10^-5

Solution

We know that χ = I/ H

I = χ H

= -0.8 x 10^-5 × 10⁶

Magnetisation in copper I= -8 Am^-1

μ_r = 1 + χ = 1 + (-0.8 × 10^-5)

=1- 10.8 x 10^-5

μ_r= 0.999

= μ H

=μ_rμ₀H

== 0.999 × 4π × 10^-7 × 10⁶

B = 0.999 × 4 × 3.14 × 10^-7 × 10⁶

Magnetic flux density B = 1.26 wb m^-2

 

Problem 2.13

A magnetic field of 1800 ampere/metre produces a magnetic flux of 3 x 10^-5 weber in an iron bar of cross sectional area 0.2 cm². Calculate magnetic permeability. (A.U. May 2015)

Given data

Magnetizing field H = 1800 Am^-1

Magnetic flux ϕ= 3 x 10^-5 weber

Area of cross section A = 0.2 cm² = 0.2 x 10^-4 m²

Solution

Magnetic flux density B= Φ/ A

Substituting the given values, we have

B = 3 × 10^-5 / 0.2 × 10^-4

B = 1.5 weber / metre²

Magnetic permeability 'µ' =B/ H

μ = 1.5/1800

Magnetic permeability μ = 8.333 x 10^-4 Hm^-1

 

Problem 2.14

The saturation magnetic induction of nickel is 0.65 weber / metre². If the density of nickel is 8906 kg/m³ and atomic weight is 58.7, calculate the magnetic moment of the nickel atom in Bohr magneton.  (A.U. Dec 2016)

Given data

Magnetic induction of nickel B = 0.65 wb m^-2

Density of Nickel ρ = 8906 kg m^-3

Atomic weight (M) = 58.7

μ₀ = 4π × 10^-7 Hm^-1

Avagadro's number N = 6.023 x 10²⁶

Solution

We know that B = Nμ₀µ_m

N = ρN/M

N is the number of atoms per unit volume (atoms/m³)

Substituting the given values, we have

We know that 1 Bohr magneton

= 9.27 × 10^-24Am²

µ_m = 5.66 × 10^-24 / 9.27 × 10^-24

μ_m = 0.61 Bohr magneton.

 

Problem 2.15

A paramagnetic material has bcc structure with a cube edge of 2.5 Å. If the saturation value of magnetisation is 1.8 x 10⁶ ampere / metre. Calculate the average magnetisation contributed per atom in Bohr magneton. (A.U. Dec. 2016)

Given data

Interatomic distance a = 2.5 Å = 2.5 ×10^-10m

Magnetisation M = 1.8 x 10⁶ Am^-1

Electronic charge 'e' = 1.6 × 10^-19coulomb

Planck's constant h = 6.625 x 10^-34Js

Solution

Number of atoms per unit volume

= No. of atoms in an unit cell / volume of the unit cell (a³)

= 2 /(2.5 × 10^-10) ³ = 1.28 × 10²⁹ m³

Total magnetisation M = 1.8 × 10⁶ Am^-1

Average magnetisation produced per atom

= 1.8 × 10⁶ / 1.28 × 10²⁹=1.406 × 10^-23Am^-1

Bohr magneton μ_{B =} eh/4πm

= 9.27 × 10^{- 24}Am^-2

Average magnetisation produced per atom in Bohr magneton=1.4065 x 10^-23 / 9.27 × 10^-24

= 1.52 Bohr magneton