## Magnetic Materials

### Maharashtra Board-Class-12th-Physics-Chapter-11

### Solutions

**Question 1. Choose the correct option.**

**(i) Intensity of magnetic field of the earth at the point inside a hollow iron box is.**

**(A) less than that outside**

**(B) more than that outside**

**(C) same as that outside**

**(D) zero**

(D) zero

**(ii) Soft iron is used to make the core of transformer because of its**

**(A) low coercivity and low retentivity**

**(B) low coercivity and high retentivity**

**(C) high coercivity and high retentivity**

**(D) high coercivity and low retentivity**

(A) low coercivity and low retentivity

**(iii) Which of the following statements is correct for diamagnetic materials?**

**(A) μ_{r} < 1**

**(B) ****χ** **is negative and low**

**(C) ****χ** **does not depend on temperature**

**(D) All of above**

(D) All of above

**(iv) A rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely. Its period of oscillation is T****′****, the ratio of T****′****/T is…..**

**(A) ** ** \(\frac{1}{2\sqrt{2}}\)**

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

**(C) 2 **

**(D) \(\frac{1}{4}\)**

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

**(v) A magnetising field of 360 Am ^{-1} produces a magnetic flux density (B) = 0.6 T in a ferromagnetic material. What is its permeability in TmA^{-1} ?**

**(A) 1/300 **

**(B) 300**

**(C) 1/600 **

**(D) 600**

(C) 1/600

**Question 2 Answer in brief.**

**(i) Which property of soft iron makes it useful for preparing electromagnet?**

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.

**(ii) What happens to a ferromagnetic material when its temperature increases above curie temperature?**

- 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.

**(iii) What should be retentivity and coercivity of permanent magnet?**

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.

**(iv) Discuss the Curie law for paramagnetic material.**

**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.

**(v) Obtain and expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.**

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

M_{0} = current x area of the loop = I(πr^{2}) = \(\frac{ev}{2πr}×πr^2\)=\(\frac{1}{2}evr\) ……..(3)

Multiplying and dividing the right hand side of the above expression by the electron mass m_{e}

M_{0} =\(\frac{e}{2m_e}(m_evr)\) = \(\frac{e}{2m_e}L_0\) …..(4)

where L_{0} = m_{e}vr 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,

L_{0} = m_{e}*v *r = \(\frac{nh}{2π}\)

Substituting for L_{0} in Eq. (4),

M_{0} = \(\frac{enh}{4πm_e}\)

For n = 1 M_{0} = \(\frac{enh}{4πm_e}\)

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-m^{2}) = 5.788 x 10^{−}^{5} eV/T.

**(vi) What does the hysteresis loop represents?**

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.

**(vii) Explain one application of electromagnet.**

**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.

**Question 3.**

**When a plate of magnetic material of size 10 cm ****× ****0.5 cm ****× ****0.2 cm (length , breadth and thickness respectively) is located in magnetising field of 0.5 ****× ****10 ^{4} Am^{-1} then a magnetic moment of 5 Am^{2} is induced in it. Find out magnetic induction in rod.**

**Given : ***l* = 10 cm, b = 0.5 cm, h = 0.2 cm, H = 0.5 × 10^{4} Am^{-1}, M = 5 A.m^{2}

The volume of the plate,

V = 10 × 0.5 × 0.2 = 1cm^{2} = 10^{-6} m^{2}

B = μ_{0}(H + M_{z}) = μ_{0 }\((H+\frac{M}{V})\)

**The magnetic induction in the plate,**

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

**= 6.290 T**

**Question 4.**

**A rod of magnetic material of cross section 0.25 cm ^{2} is located in 4000 A.m^{-1} magnetising field. Magnetic flux passing through the rod is 25 **

**×**

**10**

^{−}

^{6}**Wb. Find out**

**(a) relative permeability **

**(b) magnetic susceptibility and **

**(c) magnetisation of the rod**

**Given : **A = 0.25 cm^{2} = 25 x 10^{-6} m^{2}, H = 4000 A.m^{-1}, ø = 25 x 10^{−}^{6} Wb

Magnetic induction is

B = \(\frac{Φ}{A}\) = 1 Wb/m^{2}

(a) Β = μ_{0}μ_{r}Η

∴ 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_{m}H = 198 x 4000 = **7.92 × 10 ^{5} A/m**

**Question 5.**

**The work done for rotating a magnet with magnetic dipole moment m, through 90****° ****from its magnetic meridian is n times the work done to rotate it through 60****°****. Find the value of n.**

**Given : **θ_{0} = 0°, θ_{1} = 90°, θ_{2} = 60°, W_{1} = nW_{2}

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

W = MB (cos θ_{0} - cos θ)

∴ W_{1} = MB (cos θ_{0} - cos θ_{1})

= MB (cos 0° - cos 90°)

= MB (1 - 0)

= MB

∴ W_{2} = MB (cos 0° - cos 60°)

** **= MB

** = 0.5 MB**

∴ W_{1} = 2W_{2} = MB

Given W_{1} = nW_{2} . Therefore** n = 2**

**Question 6.**

**An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 ****× ****10**^{−}^{11}** m, with a speed of 2 ****× ****10 ^{6} m/s. Find the resultant orbital magnetic moment and angular momentum of electron. (charge on electron e = 1.6 **

**×**

**10**

^{−}

^{19}**C, mass of electron**

*m*= 9.1**×**

**10**

^{−}

^{31}**kg.)**

**Given :** r = 5.3 x 10^{−}^{11} m, v = 2 x 10^{6} 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

M_{0} = \(\frac{1}{2}\)evr

= \(\frac{1}{2}\)(1.6 x 10^{−}^{19}) (2 x 10^{6}) (5.3 × 10^{−}^{11})

= **8.48 x 10**^{−}^{24}** A.m ^{2}**

The angular momentum of the electron is

L_{0} = m_{e}vr

= (9.1 × 10^{−}^{31})(2 × 10^{6})( 5.3 × 10^{−}^{11})

= 96.46 × 10^{−}^{36} = **9.646 × 10**^{−}^{35 }**kg-m ^{2}/s**

**Question 7.**

**A paramagnetic gas has 2.0 ****× ****10 ^{26} atoms/m with atomic magnetic dipole moment of 1.5 **

**×**

**10**

^{−}

^{23}**A m2 each. The gas is at 27**

**°**

**C. (a) Find the maximum magnetization intensity of this sample. (b) If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why?**

** (Hint: Find the ratio of Thermal energy of atom of a gas ( 3/2 kBT) and maximum potential energy of the atom (mB) and draw your conclusion)**

**Given : **\(\frac{N}{V}\) = 2.0 × 10^{26} atoms/m^{3},

μ = 1.5 × 10^{-23} A.m^{2}, 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 × 10^{26})(1.5 × 10^{-23}) = =3 × 10^{3} 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_{B}T

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.

**Question 8.**

**A magnetic needle placed in uniform magnetic field has magnetic moment of 2 ****× ****10 ^{-2} A.m^{2}, and moment of inertia of 7.2 **

**×**

**10**

^{-7}kg m^{2}. It performs 10 complete oscillations in 6 s. What is the magnitude of the magnetic field ?**Given : **M = 2 × 10^{−}^{2} A.m^{2}, I = 7.2 × 10^{−}^{7} kg m^{2}, 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**^{−}^{3}**T** **= 3.943 mT**

**Question 9.**

**A short bar magnet is placed in an external magnetic field of 700 guass. When its axis makes an angle of 30****° ****with the external magnetic field, it experiences a torque of 0.014 Nm. Find the magnetic moment of the magnet, and the work done in moving it from its most stable to most unstable position.**

**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.m^{2}

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 θ_{0} **−** cos θ)

= MB (cos 0° **−** cos 180°)

= MB [1 **−** (**−**1)]

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

= 0.056 J

This the required work done.

**Question 10.**

**A magnetic needle is suspended freely so that it can rotate freely in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.2 g is kept on one end of the needle. If the pole strength of this needle is 20 Am , find the value of the vertical component of the earth's magnetic field. (g = 9.8 m/s ^{2})**

**Given** : M = 0.2 g = 2 × 10^{−}^{4} kg, q_{m} = 20 A·m, g = 9.8 m/s^{2}

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**

**Question 11. **

**The susceptibility of a paramagnetic material is ****χ** **at 27****° ****C. At what temperature its susceptibility be ****χ****/3 ?**

**Given** : χ_{m1} = χ T_{1} = 27 ^{0}C = 300 K, χ_{m2} =

By Curie's law,

M_{z} = \(C\frac{B_0}{T}\)

Since M_{z} = χ_{m}H and B_{0} = μ_{0}H

χ_{m}H = \(C\frac{μ_0H}{T}\)

∴ χ_{m} = \(C\frac{μ_0}{T}\)

∴ χ_{m }∝ \(\frac{1}{T}\)

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

∴ T_{2} = \(\frac{χ_{m1}}{χ_{m2}}\)×T_{1} = 3\(\frac{χ}{χ}\) x 300 = 900 K **= 627°C**

This gives the required temperature.

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