Solutions-Class-12-Physics-Chapter-11-Magnetic Materials-MSBSHSE

(D) zero

(A) low coercivity and low retentivity

(D) All of above

(B) \(\frac{1}{2}\)

(C) 1/600

An electromagnet should become magnetic when a current is passed through its coil but should lose its magnetism once the current is switched off. Hence, the ferromagnetic core (usually iron-based) used for an electromagnet should have high permeability and low retentivity, i.e., it should be magnetically 'sofť.

• A ferromagnetic substance is made up of tiny sections called domains. Within each domain, the atomic magnetic moments of nearest-neighbor atoms interact strongly via exchange interaction, a quantum mechanical process, and align parallel to one another even in the absence of an external magnetic field. As a result, a domain becomes magnetically saturated on its own.
• The material preserves its domain structure only at a specific temperature. Heating causes greater thermal agitation, which acts against spontaneous domain magnetization. Finally, at a crucial temperature known as the Curie point or Curie temperature, thermal agitation overcomes exchange forces, allowing the atomic magnetic moments to remain randomly oriented. So, above the Curie point, the material becomes paramagnetic. The shift from ferromagnetic to paramagnetic is characterized by an increase in disorder. When cooled below the Curie point, the material regains its ferromagnetic properties.

A permanent magnet should have a large zero-field magnetization and should need a very large reverse field to demagnetize. In other words, it should have a very broad hysteresis loop with high retentivity and very high coercivity.

Curie’s law : The magnetization of a paramagnetic material is directly proportional to the external magnetic field and inversely proportional to the absolute temperature of the material.

If a paramagnetic material at an absolute temperature T is placed in an external magnetic field of induction  the magnitude of its magnetization

M_z ∝ B_ext/T,  ∴ M_z = \(c\frac{B_{ext}}{T}\)

where the proportionality constant C is called the Curie constant.

In the Bohr model of a hydrogen atom, the electron of charge —e performs a uniform circular motion around the positively charged nucleus.

Let r, v and T be the orbital radius, speed and period of motion of the electron. Then,

T = 2πr/v  …….(1)

Therefore, the orbital magnetic moment associated with this orbital current loop has a magnitude,

I = e/T = \(\frac{ev}{2πr}\)

Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude

M₀ = current x area of the loop = I(πr²) = \(\frac{ev}{2πr}×πr^2\)=\(\frac{1}{2}evr\)     ……..(3)

Multiplying and dividing the right hand side of the above expression by the electron mass m_e

M₀ =\(\frac{e}{2m_e}(m_evr)\) = \(\frac{e}{2m_e}L_0\)   …..(4)

where L₀ = m_evr is the magnitude of the orbital angular momentum of the electron. \(\vec{M_0}\) is opposite to \(\vec{L_0}\)

∴ \(\vec{M_0}\) = \(\frac{e}{2m_e}\vec{L_0}\) …..(5)

which is the required expression

According to Bohr’s second postulate of stationary orbits in his theory of hydrogen atom, the angular momentum of the electron in the nth stationary orbit is equal to \(n\frac{h}{2π}\) , where h is the Planck constant and n is a positive integer. Thus, for an orbital electron,

L₀ = m_ev r = \(\frac{nh}{2π}\)

Substituting for L₀ in Eq. (4),

M₀ = \(\frac{enh}{4πm_e}\)

For n = 1 M₀ = \(\frac{enh}{4πm_e}\)

The quantity \(\frac{enh}{4πm_e}\) is a fundamental constant called the Bohr magneton, μ_B.

μ_B = 9.274 x 10⁻²⁴ I/T (or A-m²) = 5.788 x 10⁻⁵ eV/T.

A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

Applications of an electromagnet :

• Electromagnets are used in electric bells, loud speakers and circuit breakers.
• Large electromagnets are used in junkyard cranes and industrial cranes to lift iron scraps.
• Superconducting electromagnets are used in MRI and NMR machines, as well as in particle accelerators of cyclotron family.
• Electromagnets are used in data storage devices such as computer hard disks and magnetic tapes.

Given : l = 10 cm, b = 0.5 cm, h = 0.2 cm, H = 0.5 × 10⁴ Am^-1, M = 5 A.m²

The volume of the plate,

V = 10 × 0.5 × 0.2 = 1cm² = 10^-6 m²

B = μ₀(H + M_z) = μ₀ \((H+\frac{M}{V})\)

The magnetic induction in the plate,

∴ B = 4π x 10^-7 \((0.5×10^4+\frac{5}{10^{-6}})\)

     = 6.290 T

Given : A = 0.25 cm² = 25 x 10^-6 m², H = 4000 A.m^-1, ø = 25 x 10⁻⁶ Wb

Magnetic induction is

B = \(\frac{Φ}{A}\) = 1 Wb/m²

(a) Β = μ₀μ_rΗ

∴ The relative permeability of the material,

μ_r = \(\frac{B}{μ_0H}\) = \(\frac{1}{4×3.142×10^{-7}×4000}\) = 198.9 = 199

(b) μ_r = 1 + χ_m

∴ The magnetic susceptibility of the material,

χ_m = μ_r – 1 = 199 – 1 = 198

(c) χ_m = \(\frac{M_z}{H}\) 

The magnetization of the rod,

Mz = χ_mH = 198 x 4000 = 7.92 × 10⁵ A/m

Given : θ₀ = 0°, θ₁ = 90°, θ₂ = 60°, W₁ = nW₂

The work done by an external agent to rotate the magnet from 00 to 0 is

W = MB (cos θ₀ - cos θ)

∴ W₁ = MB (cos θ₀ - cos θ₁)

        = MB (cos 0° - cos 90°)

        = MB (1 - 0)

        = MB

∴ W₂ = MB (cos 0° - cos 60°)

        = MB  

        = 0.5 MB

∴ W₁ = 2W₂ = MB

Given W₁ = nW₂ . Therefore n = 2

Given : r = 5.3 x 10⁻¹¹ m, v = 2 x 10⁶ m/s, e = 1.6 x 10⁻¹⁹ C, m_e = 9.1 x 10⁻³¹ kg

The orbital magnetic moment of the electron is

M₀ = \(\frac{1}{2}\)evr

= \(\frac{1}{2}\)(1.6 x 10⁻¹⁹) (2 x 10⁶) (5.3 × 10⁻¹¹)

= 8.48 x 10⁻²⁴ A.m²

The angular momentum of the electron is

L₀ = m_evr

    = (9.1 × 10⁻³¹)(2 × 10⁶)( 5.3 × 10⁻¹¹)

    = 96.46 × 10⁻³⁶ = 9.646 × 10⁻³⁵ kg-m²/s

Given : \(\frac{N}{V}\) = 2.0 × 10²⁶ atoms/m³,

μ = 1.5 × 10^-23 A.m², T = 27 + 273 = 300 K, B = 3 T, k_B = 1.38 × 10^-23 J/K,

1 eV = 1.6 × 10^-19 J

(a) The maximum magnetization of the material,

M_Z = \(\frac{N}{V}\)μ = (2.0 × 10²⁶)(1.5 × 10^-23) = =3 × 10³ A/m

(b) The maximum orientation energy per atom is

U_max = −μB cos 180° = μB

= (1.5 × 10^-23)(3) = 4.5 × 10^-23 J =  eV = 2.8 × 10^-4 eV

The average thermal energy of each atom,

E = \(\frac{3}{2}\)k_BT

where k_B is the Botzmann constant.

∴ E = 1.5(1.38 x 10⁻²³)(300)

     = 6.21 x 10⁻²¹ J = \(\frac{6.21×10^{-21}}{1.6×10^{-19}}\)  eV = 3.9 × 10⁻² eV

Since the thermal energy of randomization is about two orders of magnitude greater than the magnetic potential energy of orientation, saturation magnetization will not be achieved at 300 K.

Given : M = 2 × 10⁻² A.m², I = 7.2 × 10⁻⁷ kg m², T = 6/10 = 0.6 s

T = 2π\(\sqrt{\frac{I}{MB}}\)

The magnitude of the magnetic field is

B = \(\frac{4π^2I}{MT^2}\) = \(\frac{(4)(3.14)^2(7.2×10^{-7}}{(2×10^{-2})(0.6)^2}\) = 3.943 × 10⁻³T = 3.943 mT

Given : B = 700 gauss = 0.07 tesla, θ = 30°, τ = 0.014 N·m,

τ = MB sin θ

The magnetic moment of the magnet is

M = \(\frac{τ}{B\,sin\,θ}=\frac{0.014}{(0.07)(sin\,30^0)}\) = 0.4 A.m²

The most stable state of the bar magnet is for θ = 0°.

It is in the most unstable state when θ = 180°.

Thus, the work done in moving the bar magnet from 0° to 180° is

W = MB (cos θ₀ − cos θ)

= MB (cos 0° − cos 180°)

= MB [1 − (−1)]

= 2MB = (2)(0.4)(0.07)

= 0.056 J

This the required work done.

Given : M = 0.2 g = 2 × 10⁻⁴ kg, q_m = 20 A·m, g = 9.8 m/s²

Without the added weight at one end, the needle will dip in the direction of the resultant magnetic field inclined with the horizontal. The torque due to the added weight about the vertical axis through the centre balances the torque of the couple due to the vertical component of the Earth's magnetic field.

∴ (Mg)\(\frac{L}{2}\) = (q_m B_v)L

The vertical component of the Earth's magnetic field,

B_v = \(\frac{Mg}{2q_m}\) = \(\frac{2×10^{-4}×9.8}{2×20}\) = 4.9 × 10⁻⁵ T

Given : χ_m1 = χ  T₁ = 27 ⁰C = 300 K, χ_m2 =

By Curie's law,

M_z = \(C\frac{B_0}{T}\)

Since M_z = χ_mH and B₀ = μ₀H

χ_mH = \(C\frac{μ_0H}{T}\)

∴ χ_m = \(C\frac{μ_0}{T}\)

∴ χ_m ∝ \(\frac{1}{T}\)

∴ \(\frac{χ_{m1}}{χ_{m2}}\) = \(\frac{1}{T}\)

∴ T₂ = \(\frac{χ_{m1}}{χ_{m2}}\)×T₁ = 3\(\frac{χ}{χ}\) x 300 = 900 K = 627°C

This gives the required temperature.