SSP 2 ASSIGNMENT 5

solution to ASSIGNMENT 5 OF SSP 2

1. A magnetic material has a magnetization of 3300 amperes/meter and flux density of 0.0044
   weber/meter 2 . Calculate the magnetic force and relative permeability of the material.

Ans : To calculate the magnetic force and relative permeability, we need to use the following formulas:

1. Magnetic field strength (H) = Magnetization (M) / μ₀ (where μ₀ is the permeability of free space, approximately 4π x 10^-7 H/m)

H = 3300 A/m / (4π x 10^-7 H/m) = 2.62 x 10^6 A/m

1. Magnetic force (F) = Magnetic field strength (H) x Flux density (B)

F = 2.62 x 10^6 A/m x 0.0044 Wb/m^2 = 11.53 N

1. Relative permeability (μr) = Flux density (B) / (Magnetic field strength (H) x μ₀)

μr = 0.0044 Wb/m^2 / (2.62 x 10^6 A/m x 4π x 10^-7 H/m) = 133.55

So, the magnetic force is approximately 11.53 Newtons, and the relative permeability is approximately 133.55.

Note: The unit of magnetization (M) is Amperes per meter (A/m), and the unit of flux density (B) is Teslas (T) or Webers per square meter (Wb/m^2). In this problem, the given flux density is in Wb/m^2, which is equivalent to Teslas (T).

2.A typical magnetic field achievable with an electromagnet with iron core is about 1 Tesla. Compare the magnetic interaction energy µBB of an electron spin magnetic dipole moment with kBT at room temperature (300K) and show that at ordinary temperatures the approximation KBT/µBB >> 1 is valid

Ans :

The magnetic interaction energy between two magnetic dipole moments, such as an electron spin and an external magnetic field, is given by:

µB = (eh/2me) = 9.274 × 10^-24 J/T (Bohr magneton)

B = 1 Tesla (typical magnetic field strength with an iron core electromagnet)

k_B = 1.381 × 10^-23 J/K (Boltzmann constant)

T = 300 K (room temperature)

Now, let’s calculate:

µB × B = 9.274 × 10^-24 J/T × 1 T = 9.274 × 10^-24 J

k_B × T = 1.381 × 10^-23 J/K × 300 K = 4.143 × 10^-21 J

Now, let’s compare:

k_B × T / (µB × B) = 4.143 × 10^-21 J / (9.274 × 10^-24 J) ≈ 446 >> 1

Therefore, at ordinary temperatures, the approximation k_B × T / µB × B >> 1 is indeed valid. This means that thermal fluctuations dominate over magnetic interactions at room temperature.

3.The saturation magnetic induction of nickel is 0.65Wb/m2
. If the density of nickel is 8906Kg/m3 and its atomic weight is 58.7, Calculate the magnetic moment of nickel atom in Bohr magneton.

Ans:

1. First, we need to find the number of atoms per unit volume (n) using the density and atomic weight:

n = density / (atomic weight x Avogadro’s constant)
= 8906 kg/m³ / (58.7 g/mol x 6.022 x 10²³ mol⁻¹)
= 2.54 x 10²⁸ atoms/m³

1. Next, we can find the magnetic moment per unit volume (M) using the saturation magnetic induction (Bs):

M = Bs / μ₀
= 0.65 T x 4π x 10⁻⁷ T·m³/A
= 8.17 x 10⁴ A/m

1. Now, we can calculate the magnetic moment per atom (m):

m = M / n
= 8.17 x 10⁴ A/m / 2.54 x 10²⁸ atoms/m³
= 3.21 x 10⁻²⁴ A·m²/atom

1. Finally, we can express this magnetic moment in terms of the Bohr magneton (μB):

μB = 9.274 x 10⁻²⁴ J/T
m / μB = 3.21 x 10⁻²⁴ A·m²/atom / 9.274 x 10⁻²⁴ J/T
≈ 0.346 μB/atom

So, the magnetic moment of a nickel atom is approximately 0.346 Bohr magnetons.

4. A solenoid 0.5 meter long and containing 500 turns of wire is passed by a current of 10 amperes. When placed in an atmosphere of oxygen at t= -150C, it exhibits an increase of magnetic induction of B =1.04 x10-8 wb/m2 above the measured in vacuum. Compute the magnetic susceptibility of oxygen at -150C and magnetic dipole moment of oxygen.

Ans:

1. First, we need to find the magnetic field strength (H) inside the solenoid:

H = NI / L
= 500 turns x 10 A / 0.5 m
= 1000 A/m

1. Next, we can find the magnetic induction (B) inside the solenoid using the magnetic permeability of free space (μ₀):

B = μ₀ x H
= 4π x 10⁻⁷ T·m/A x 1000 A/m
= 0.004π T

1. The increase in magnetic induction (ΔB) is given as 1.04 x 10⁻⁸ Wb/m². We can find the magnetic susceptibility (χ) using:

χ = ΔB / (μ₀ x H)
= 1.04 x 10⁻⁸ Wb/m² / (4π x 10⁻⁷ T·m/A x 1000 A/m)
≈ 8.25 x 10⁻⁶

1. Now, we can find the magnetic dipole moment (m) using the magnetic susceptibility and the number of molecules per unit volume (n):

m = χ x H / (μ₀ x n)

We need to find the number of molecules per unit volume (n). Let’s assume the density of oxygen at -150°C is approximately 1.426 kg/m³ (using the ideal gas law and the molar mass of oxygen).

n = density / (molar mass x Avogadro’s constant)
= 1.426 kg/m³ / (32 g/mol x 6.022 x 10²³ mol⁻¹)
≈ 7.39 x 10²⁵ molecules/m³

Now, we can calculate the magnetic dipole moment:

m = 8.25 x 10⁻⁶ x 1000 A/m / (4π x 10⁻⁷ T·m/A x 7.39 x 10²⁵ molecules/m³)
≈ 9.33 x 10⁻²⁹ A·m²

So, the magnetic susceptibility of oxygen at -150°C is approximately 8.25 x 10⁻⁶, and the magnetic dipole moment of oxygen is approximately 9.33 x 10⁻²⁹ A·m².

5.An atom contains 10 electrons revolving in a circular path of radius 10-11 meter. Assuming
homogenous charge distribution, calculate the orbital dipole moment of the model in Bohr magneton

Ans :

1. First, we need to find the angular momentum (L) of the electrons:

L = m x v x r

where m is the mass of an electron (approximately 9.11 x 10^-31 kg), v is the velocity of the electron (which we can find using the Bohr model), and r is the radius of the circular path (given as 10^-11 m).

Using the Bohr model, we can find the velocity:

v = h / (2 x π x m x r)
= 6.626 x 10^-34 J·s / (2 x π x 9.11 x 10^-31 kg x 10^-11 m)
≈ 1.15 x 10^6 m/s

Now, we can calculate the angular momentum:

L = 9.11 x 10^-31 kg x 1.15 x 10^6 m/s x 10^-11 m
≈ 1.05 x 10^-24 J·s

1. Next, we can find the orbital dipole moment (μ) using:

μ = L / (2 x e)

where e is the elementary charge (approximately 1.602 x 10^-19 C):

μ = 1.05 x 10^-24 J·s / (2 x 1.602 x 10^-19 C)
≈ 3.28 x 10^-24 A·m

1. Finally, we can express this orbital dipole moment in terms of the Bohr magneton (μB):

μB = 9.274 x 10^-24 J/T

μ / μB ≈ 3.28 x 10^-24 A·m / 9.274 x 10^-24 J/T
≈ 0.353 μB

So, the orbital dipole moment of the model is approximately 0.353 Bohr magnetons.

6. Consider a helium atom in its ground sate (1s). the mean radius in the Langevin formula may be approximated by the Bohr-radius, 0.528nm. Density if helium is 0.178 Kg/m3 . Calculate the
diamagnetic susceptibility of a helium atom.

Ans :

1. First, we need to find the number of atoms per unit volume (n) using the density and atomic weight:

n = density / (atomic weight x Avogadro’s constant)
= 0.178 kg/m³ / (4.0026 g/mol x 6.022 x 10²³ mol⁻¹)
≈ 7.26 x 10²⁵ atoms/m³

1. Next, we can find the diamagnetic susceptibility (χ) using the Langevin formula:

χ = -(μ₀ x e² x r²) / (6 x m x k_B x T)

where:

• μ₀ is the magnetic permeability of free space (4π x 10⁻⁷ T·m/A)
• e is the elementary charge (1.602 x 10⁻¹⁹ C)
• r is the mean radius (approximated by the Bohr radius, 0.528 x 10⁻¹⁰ m)
• m is the mass of an electron (9.11 x 10⁻³¹ kg)
• k_B is the Boltzmann constant (1.381 x 10⁻²³ J/K)
• T is the temperature (assuming room temperature, 300 K)

χ = -(4π x 10⁻⁷ T·m/A x (1.602 x 10⁻¹⁹ C)² x (0.528 x 10⁻¹⁰ m)²) / (6 x 9.11 x 10⁻³¹ kg x 1.381 x 10⁻²³ J/K x 300 K)
≈ -2.44 x 10⁻⁶

1. Finally, we can express this susceptibility per atom (χ_atom) using:

χ_atom = χ / n
= -2.44 x 10⁻⁶ / 7.26 x 10²⁵ atoms/m³
≈ -3.36 x 10⁻³² m³/atom

So, the diamagnetic susceptibility of a helium atom is approximately -3.36 x 10⁻³² m³/atom.