Magnetic Materials: Important Solved Problems

SOLVED PROBLEMS

ANNA UNIVERSITY SOLVED PROBLEMS

1. The saturation magnetic induction of nickel is 0.65 Wb/m². 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.

Solution:

Given B_S = 0.65 Wb/m²

 ρ = 89.6 kg/m³

Atomic weight = 58.7

We know B_S = Nμ₀μ_m (or) μ_m = B_S / Nμ₀

We know μ₀ = 4 × 10^-7

N = ρA / Atomic weight

Here N is the number atoms/m³ in 1 kg of substance

 N = 8906×6.023×10²⁶  /  58.7

 ∴ N=9.14 × 10²⁸ atoms/m³

μ_m = 0.65 / [ 9.14 × 10^‒28 × 4π × 10^‒7 ]

= 5.66 × 10^‒24 Am²

We know 1 Bohr Magneton = 9.27 x 10^‒24 Am²

μ_m = 5.66×10^‒24 /  9.27×10^‒24  Bohr mageton

= 0.61 Bohr mageton

H=0.61 μB

2. The rare earth element gadolinium is ferromagnetic below 16°C with 7.1 Bohr magneton per atom. Calculate the magnetic moment per gram. What is the value of saturation magnetisation, given that the atomic weight of gadolinium is 157.26 and its density is 7.8 x 10³ kg/m³.

Solution:

Number of atoms per kg = [ 6.025 × 10²⁶ × 7.8 × 10³ ] / 157.26

= 2.9883 × 10²⁸

∴ Number of atoms per gram = 2.9883 × 10²⁵

(i) Magnetic moment per gram = 2.9883 × 10²⁵ × 7.1 Bohr magneton

Since 1 Bohr magneton = 9.27 × 10^‒24 Am²

Magnetic moment per gram = 2.9883 × 10²⁵ × 7.1 × 9.27 × 10^‒24 Am²

 m = 1966.809 Am²

 (ii) Saturation magnetisation B_S = Nμ₀μ_m

 = 2.9883 × 10²⁸ × 4π × 10^‒7 × 9.27 × 10^‒24

B_S = 0.3481 Wbm^‒2

3. A paramagnetic material has a magnetic field intensity of 10⁴ A/m. If the susceptibility of the material at room temperature is 3.7 × 10^‒3 calculate the magnetization and flux density in the material.

Solution:

(i) We know susceptibility χ = I / H

 ∴ Intensity of magnetisation I = χH

 = 3.7 × 10^‒3 × 10⁴

 I = 37 Am^‒1

(ii) Flux density B = μ₀ [H+I]

В = 4π × 10^‒7 [10⁴ +37]

B = 0.012612 Wb/m²

4. A magnetic field strength of 2× 10⁵ amperes/m is applied to a paramagnetic material with a relative permeability of 1.01. Calculate the values of B and M.

Solution:

Given data:

Magnetic field strength (H) = 2 × 10⁵ Amp/m

Relative permeability (μ_r) = 1.01

Permeability in free space (μ₀) = 4л× 10^‒7

Formula:

(i) We know magnetic permeability = μ = B/H

(or)

μ₀μ_r = B/H    [ μ= μ₀μ_r]

Magnetic flux density B = μ₀μ_rH

∴ B=4×3.14 × 10^‒7 × 1.01 × 2 × 10⁵

∴ B=0.2537 weber/m²

∴ Magnetic flux density (B) = 0.2537 Wb/m²

(ii) We know magnetic susceptibility χ_m = μ_r ‒ 1

 (or) M/H = μ_r ‒ 1  [χ_m=M/H]

Intensity of magnetisation M = H (μ_r ‒ 1)

 ∴ M=2×10⁵ (1.01‒1)

∴ M=2000 Amp/m

Intensity of magnetisation (M) = 2000 Amp/m

5. Magnetic field intensity of a paramagnetic material is 10⁴ A/m. At room temperature, its susceptibility is 3.7 × 10^‒3. Calculate the magnetization in the material.

Solution:

We know χ = I/H

(or) I = Hχ

I = 3.7 × 10^‒3 × 10⁴

I = 37 Am^‒1

∴ Magnetization in the material = 37 Am^‒1

6. A magnetic field of 2000 A/m is applied to a material which has a susceptibility of 1000. Calculate (i) Intensity of magnetization and (ii) Flux density.

Solution

Given Data:

H = 2000 A/m

χ=1000

I=?

B=?

Formulae:

(i) Intensity of Magnetisation I = χ_mH

 I = 1000 × 2000

 I = 2×10⁶ A/m

∴ Intensity of magnetization I = 2 × 10⁶ A/m

(ii) Flux density

 B = μH

Here μ = 1+χ_m

μ = 1+1000

μ= 1001

∴ B=1001 × 2000 × 4π × 10^‒7

∴ B= 2.5145 Wb/m²

∴ Flux density B = 2.5145 Wb/m²

7. The magnetic field strength of Silicon is 1500 A/m. If the magnetic susceptibility is –0.3 × 10^‒5, calculate the magnetization and flux density in Silicon.

Solution

Given Data:

 H = 1500 A/m

χ_m= ‒0.3 × 10^‒5

Formulae:

(i) Intensity of magnetization I = χ_m H

∴ I = ‒0.3 × 10^‒5 × 1500

∴ I = ‒4.5 × 10^‒3 A/m

 ∴ Intensity of magnetization I = −4.5 × 10^‒3 A/m

(ii) Flux density B = μ H

 B = (1+χ_m) H

B = (1 − 0.3 × 10^‒5) × 1500

B = 0.999997 × 1500

 B = 1499.9 Wh/m²

Magnetic flux density B = 1499.9 Wb/m²

ADDITIONAL SOLVED PROBLEMS

1. The magnetic field strength in copper is 10⁶ A/m. If the magnetic susceptibility of copper is ‒0.8 × 10^‒5, calculate the flux density and magentisation in copper.

Solution

χ_m= I/H

I = χ_mH = ‒0.8 × 10^‒5 × 10⁶

I = ‒8 A/m

μ_r = 1+ χ_m

= 1‒ 0.8×10^‒5

μ_r = 0.999992

B = μH = 0.999992 × 4π × 10^‒7 × 10⁶

B = 1.26 wb/m²

2. A piece of ferric oxide with magnetic field intensity 10⁶ A/m and susceptibility is 1.5× 10^‒3. Find the magnetisation of the material.

Solution

I = χ_mH = 1.5 × 10^‒3 × 10⁶

 I = 1500 A/m

3. A magnetic material of flux density and magnetisation are 0.0044 Wb/m² and 3300 Alm respectively. Calculate the magnetising force and relative permeability of the material.

Solution

χ_m= I/H

χ_m= Iμ / Hμ = Iμ / B

I/B = χ_m / μ  = (μ_r‒1) / μ₀μ_r

I/B = 3300 / 0.0044

= 750000

[ μ_r‒1 ] / μ_r  = 4π × 10⁻⁷ × 750000 = 0.94247

1 ‒ 1/μ_r = 0.94247

1/μ_r= 0.0575

 μ_r = 17.3

202.39 A/m

H = B/μ

= 0.0044 / [4π × 10^‒7 × 17.3]

= 202.39 A/m

 μ_r = 1+ χ_m  = 1+1.5 × 10^‒3

 μ_r =1.0015

B=μH = 1.0015 × 4 × 10⁷ × 10⁶

 B=1.259 Wh/m²

4. Calculate the magnetic moment per unit volume in copper when subjected to a field whose magnetic inside copper is 10⁴ A/m. The magnetic susceptibility of copper is ‒0.5 × 10^‒5.

Solution

Magnetic moment per unit volume = Intensity of Magnetisation (I)

χ_m= I/H

 I = χ_mH = ‒0.5 × 10^‒5 × 10⁴

I = ‒0.05 A/m

ASSIGNMENT PROBLEMS

1. The rare element gadolinium is ferromagnetic below 10°C with 7.1 Bohr magneton per atom. Calculate the magnetic moment per gram. What is the value of saturation magnetization? The atomic weight of gadolinium is 157.26 and its density is 7.8 × 10³ kg/m³.

 [Solution: (i) 2.7 × 10²² Bohr magneton (ii) 5.1 × 10⁶ A/m]

2. The saturation value of magnetization of iron is 1.76 × 10⁶ amp/m. Iron has body centered cubic structure with elementary cube edge of 2.86 AU. Calculate the average number of Bohr magneton contributed to magnetisation per atom. [Solution: 2.22 Bohr magneton/atom]

3. A paramagnetic material when subjected to homogeneous field of 10⁶ A/m at room temperature of 30°C. Find the average magnetic moment along field direction per spin in Bohr magneton. [Ans: 2.79 × 10^‒3 Bohr Magneton/spin]