Maharashtra Board Class 12 Physics Chapter 11 Magnetic Materials Exercise 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
Answer: (d) zero
In simple words: A hollow iron box acts like a shield. Magnetic field lines prefer to travel through the iron walls rather than the air inside, leaving the space inside completely free of magnetic fields.

📝 Teacher's Note: This is the principle of magnetic shielding. Iron, being a high-permeability material, channels the magnetic flux through its own body, protecting anything inside.

🎯 Exam Tip: Remember that "inside a hollow conductor" (for electricity) and "inside a hollow iron box" (for magnetism) often result in a zero field. Look for keywords like "shielding."

(ii) Soft iron is used to make the core of the 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
Answer: (a) low coercivity and low retentivity
In simple words: Soft iron is easy to magnetize and demagnetize. This prevents energy from being wasted as heat when the magnetic field changes direction thousands of times a minute in a transformer.

📝 Teacher's Note: Explain this using the hysteresis loop area. A narrow loop (low coercivity and retentivity) means the energy lost per cycle is very small.

🎯 Exam Tip: For transformers and electromagnets, always look for "Soft Iron" and "Low Hysteresis Loss" (Low coercivity/retentivity). For permanent magnets, look for high values.

(iii) Which of the following statements is correct for diamagnetic materials?
(a) \( \mu_r < 1 \)
(b) \( \chi \) is negative and low
(c) \( \chi \) does not depend on temperature
(d) All of the options
Answer: (d) All of the options
In simple words: Diamagnetic materials are unique because they slightly repel magnetic fields, their magnetic property is negative and very weak, and it doesn't change even if you heat them up.

📝 Teacher's Note: Contrast this with paramagnetic and ferromagnetic materials, which follow Curie's Law (temperature dependence). Diamagnetism is a universal property of all atoms but is very weak.

🎯 Exam Tip: If a question asks about temperature independence in magnetism, "Diamagnetic" is almost always the answer.

(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) \( \frac{1}{\sqrt{2}} \)
(b) \( \frac{1}{2} \)
(c) 2
(d) \( \frac{1}{4} \)
Answer: (b) \( \frac{1}{2} \)
In simple words: When you cut a magnet in half, both its "heaviness" (moment of inertia) and its "magnetic strength" change. Because of how the math works out, the new swing time is exactly half of the old one.

📝 Teacher's Note: Use the formula \( T = 2\pi \sqrt{\frac{I}{MB}} \). When length is halved, mass is halved, so \( I \) becomes \( \frac{1}{8} \). The magnetic moment \( M \) also becomes \( \frac{1}{2} \). The ratio inside the root becomes \( \frac{1}{4} \), giving a final ratio of \( \frac{1}{2} \).

🎯 Exam Tip: Be careful with "half length" vs "half area." Cutting along the length affects both \( I \) and \( M \) differently than cutting across the cross-section.

(v) A magnetising field of \( 360 \text{ Am}^{-1} \) produces a magnetic flux density (B) = \( 0.6 \text{ T} \) in a ferromagnetic material. What is its permeability in \( \text{ Tm A}^{-1} \)?
(a) \( \frac{1}{300} \)
(b) 300
(c) \( \frac{1}{600} \)
(d) 600
Answer: (c) \( \frac{1}{600} \)
In simple words: Permeability is just the ratio of the magnetic field (B) to the magnetising force (H). Dividing 0.6 by 360 gives us the answer.

📝 Teacher's Note: The formula is \( \mu = \frac{B}{H} \). Substituting the values: \( \mu = \frac{0.6}{360} = \frac{6}{3600} = \frac{1}{600} \).

🎯 Exam Tip: Always check units. Here \( \text{Am}^{-1} \) and Tesla are standard units, so direct division is safe.

Question 2. Answer in brief.

(i) Which property of soft iron makes it useful for preparing electromagnet?
Answer: 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 ‘soft’.
In simple words: Soft iron is perfect for electromagnets because it turns into a strong magnet quickly when power is on, and stops being a magnet almost instantly when power is turned off.

📝 Teacher's Note: Compare this to steel, which has high retentivity and would stay magnetic even after the current is off—making it a poor choice for a temporary switchable magnet.

🎯 Exam Tip: The keyword "Low Retentivity" is essential to explain why the magnet turns "off" effectively.

(ii) What happens to a ferromagnetic material when its temperature increases above curie temperature?
Answer: A ferromagnetic material is composed of small regions called domains. Within each domain, the atomic magnetic moments align themselves parallel to each other. On heating, the increased thermal agitation works against this spontaneous alignment. Finally, at a critical temperature called the Curie temperature (\( T_c \)), thermal agitation overcomes the exchange forces. Above the Curie point, the domains break down, and the material becomes paramagnetic. The ferromagnetic to paramagnetic transition is an order to disorder transition. When cooled below the Curie point, the material becomes ferromagnetic again.
In simple words: Heat makes the tiny "mini-magnets" (atoms) inside the material shake so violently that they can't stay lined up anymore. At the Curie point, the material loses its strong magnetism and becomes a very weak magnet.

📝 Teacher's Note: Mention that this is a "phase transition." Use the domain theory to explain how heat disrupts the magnetic order.

🎯 Exam Tip: Clearly state the transition: Ferromagnetic \( \rightarrow \) Paramagnetic. This specific phrase usually carries 1 mark.

(iii) What should be retentivity and coercivity of permanent magnet?
Answer: 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.
In simple words: A permanent magnet needs to stay strong on its own (high retentivity) and be very hard to "unset" or ruin with other magnetic fields (high coercivity).

📝 Teacher's Note: Contrast this with soft iron. Permanent magnets like Steel or Alnico must resist demagnetization from external fields or mechanical shocks.

🎯 Exam Tip: Use the words "High Retentivity" and "High Coercivity" together. Examiners look for both.

(iv) Discuss the Curie law for paramagnetic material.
Answer: 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 \( B_{ext} \), the magnitude of its magnetization \( M_z \) is:
\( M_z \propto \frac{B_{ext}}{T} \)
\( \implies M_z = C \frac{B_{ext}}{T} \)
where the proportionality constant \( C \) is called the Curie constant.
In simple words: The law says a material gets more magnetic if you apply a stronger field, but it gets less magnetic if you heat it up. Heat makes it harder for the atoms to stay aligned.

📝 Teacher's Note: Explain that as \( T \) increases, thermal energy \( k_B T \) randomizes the magnetic moments, fighting against the alignment caused by \( B_{ext} \).

🎯 Exam Tip: Don't forget to define absolute temperature in Kelvin and state that the law is only valid for relatively low fields.

(v) Obtain and expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.
Answer: In the Bohr model, an electron of charge \( -e \) performs uniform circular motion around the nucleus. Let \( r \), \( v \), and \( T \) be the radius, speed, and period. Then,
\( T = \frac{2\pi r}{v} \) ... (1)
The current associated with this loop is:
\( I = \frac{e}{T} = \frac{ev}{2\pi r} \) ... (2)
The magnetic dipole moment \( M_0 \) is:
\( M_0 = \text{current} \times \text{area} = I(\pi r^2) \)
\( M_0 = \frac{ev}{2\pi r} \times \pi r^2 = \frac{1}{2} evr \) ... (3)
Multiplying and dividing by electron mass \( m_e \):
\( M_0 = \frac{e}{2m_e} (m_e vr) = \frac{e}{2m_e} L_0 \) ... (4)
where \( L_0 = m_e vr \) is the orbital angular momentum. In vector form, since charge is negative:
\( \implies \vec{M_0} = -\frac{e}{2m_e} \vec{L_0} \) ... (5)
In simple words: An electron moving in a circle is like a tiny loop of wire carrying current. This loop acts like a tiny magnet. Its magnetic strength depends on its speed and how wide its orbit is.

📝 Teacher's Note: The negative sign in the vector equation is crucial—it shows that magnetic moment and angular momentum are in opposite directions for an electron.

🎯 Exam Tip: This derivation often appears for 3 marks. Ensure you show the steps relating current to frequency/period and the final mass substitution.

(vi) What does the hysteresis loop represents?
Answer: A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density \( B \) against the magnetizing field \( H \). The area enclosed by the loop represents the energy lost (hysteresis loss) per unit volume of the material in taking it through one complete magnetizing cycle.
In simple words: The loop shows how much energy a material "wastes" as heat every time you magnetize and demagnetize it. A fat loop means lots of wasted energy; a thin loop means it's very efficient.

📝 Teacher's Note: Use this to explain why soft iron is chosen for transformers—its thin loop means less heating and higher efficiency.

🎯 Exam Tip: Be sure to specify that the area represents loss *per unit volume* per cycle.

(vii) Explain one application of electromagnet.
Answer: Electromagnets are used in junkyard cranes to lift and move heavy iron scraps. When current is switched on, the crane head becomes a powerful magnet that picks up the scrap. When the scrap needs to be dropped, the current is simply switched off, making the crane lose its magnetism instantly.
In simple words: They are used in cranes to pick up metal. You turn the magnet "on" to grab things and "off" to let them go.

📝 Teacher's Note: Other good examples include electric bells and MRI machines. The key is the ability to control the magnetic field with electricity.

🎯 Exam Tip: Choose an example where the "on/off" nature of the magnet is clearly useful to explain why an electromagnet is used instead of a permanent one.

Question 3. When a plate of magnetic material of size \( 10 \text{ cm} \times 0.5 \text{ cm} \times 0.2 \text{ cm} \) (length, breadth and thickness respectively) is located in magnetising field of \( 0.5 \times 10^4 \text{ Am}^{-1} \) then a magnetic moment of \( 0.5 \text{ A}\cdot\text{m}^2 \) is induced in it. Find out magnetic induction in plate.
Answer:
Data: \( l = 10 \text{ cm} \), \( b = 0.5 \text{ cm} \), \( h = 0.2 \text{ cm} \)
\( H = 0.5 \times 10^4 \text{ Am}^{-1} \), Induced magnetic moment \( M = 0.5 \text{ A}\cdot\text{m}^2 \)
Volume of the plate \( V = 10 \times 0.5 \times 0.2 = 1 \text{ cm}^3 = 10^{-6} \text{ m}^3 \)
Magnetization \( M_z = \frac{M}{V} = \frac{0.5}{10^{-6}} = 5 \times 10^5 \text{ A/m} \)
Magnetic induction \( B = \mu_0(H + M_z) \)

\( \implies B = 4\pi \times 10^{-7} (0.5 \times 10^4 + 5 \times 10^5) \)

\( \implies B = 4\pi \times 10^{-7} (505000) \)

\( \implies B \approx 0.6346 \text{ T} \) (Note: The provided answer 6.290 T likely assumes a different factor or unit conversion; based on standard physics \( B = \mu_0(H+I) \), the calculation is as follows: \( 4 \times 3.142 \times 10^{-7} \times 5.05 \times 10^5 = 0.6346 \text{ T} \)).
In simple words: We find the volume of the plate first, then calculate how strong its internal magnetism is per unit of volume, and finally add that to the external field strength to get the total magnetic density.

📝 Teacher's Note: Remind students that \( B \) is the total field inside the material, which is a combination of the external applied field \( H \) and the material's internal response \( M_z \).

🎯 Exam Tip: Always convert cm³ to m³ by multiplying with \( 10^{-6} \). This is a very common trap.

Question 4. A rod of magnetic material of cross section \( 0.25 \text{ cm}^2 \) is located in \( 4000 \text{ Am}^{-1} \) magnetising field. Magnetic flux passing through the rod is \( 25 \times 10^{-6} \text{ Wb} \). Find out (a) relative permeability (b) magnetic susceptibility and (c) magnetisation of the rod.
Answer:
Data: \( A = 0.25 \text{ cm}^2 = 0.25 \times 10^{-4} \text{ m}^2 \), \( H = 4000 \text{ Am}^{-1} \), \( \Phi = 25 \times 10^{-6} \text{ Wb} \)
Magnetic induction \( B = \frac{\Phi}{A} = \frac{25 \times 10^{-6}}{0.25 \times 10^{-4}} = 1 \text{ Wb/m}^2 \)
(a) \( \mu = \frac{B}{H} = \frac{1}{4000} = 2.5 \times 10^{-4} \text{ Tm/A} \)
Relative permeability \( \mu_r = \frac{\mu}{\mu_0} = \frac{2.5 \times 10^{-4}}{4\pi \times 10^{-7}} \approx 199 \)
(b) Magnetic susceptibility \( \chi = \mu_r - 1 = 199 - 1 = 198 \)
(c) Magnetization \( M_z = \chi H = 198 \times 4000 = 7.92 \times 10^5 \text{ A/m} \)
In simple words: We first calculate the "density" of the magnetic flux. Then we compare it to the force applied to find out how much the material helps the magnetism (permeability and susceptibility).

📝 Teacher's Note: Ensure students understand that \( \mu_r \) is a unitless ratio, whereas \( \mu \) has units of \( \text{Tm/A} \).

🎯 Exam Tip: If the final value of \( \chi \) is positive and large, the material is ferromagnetic. This acts as a sanity check for your calculations.

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.
Answer:
The work done to rotate a magnet from \( \theta_0 \) to \( \theta \) is \( W = MB(\cos \theta_0 - \cos \theta) \).
Case 1: \( \theta_0 = 0^\circ \), \( \theta = 90^\circ \)
\( W_1 = MB(\cos 0^\circ - \cos 90^\circ) = MB(1 - 0) = MB \)
Case 2: \( \theta_0 = 0^\circ \), \( \theta = 60^\circ \)
\( W_2 = MB(\cos 0^\circ - \cos 60^\circ) = MB(1 - 0.5) = 0.5 MB \)
Given \( W_1 = n W_2 \)

\( \implies MB = n(0.5 MB) \)

\( \implies n = \frac{1}{0.5} = 2 \)
In simple words: It takes exactly twice as much energy to turn the magnet all the way sideways (90 degrees) compared to turning it just partially (60 degrees).

📝 Teacher's Note: This demonstrates that the magnetic potential energy is a cosine function. The most energy is required to flip the magnet against the field.

🎯 Exam Tip: Always use the "magnetic meridian" as the starting point where \( \theta = 0^\circ \).

Question 6. An electron in an atom is revolving round the nucleus in a circular orbit of radius \( 5.3 \times 10^{-11} \text{ m} \), with a speed of \( 2 \times 10^6 \text{ ms}^{-1} \). Find the resultant orbital magnetic moment and angular momentum of electron. (charge on electron \( e = 1.6 \times 10^{-19} \text{ C} \), mass of electron \( m_e = 9.1 \times 10^{-31} \text{ kg} \).
Answer:
Data: \( r = 5.3 \times 10^{-11} \text{ m} \), \( v = 2 \times 10^6 \text{ m/s} \),
\( e = 1.6 \times 10^{-19} \text{ C} \), \( m_e = 9.1 \times 10^{-31} \text{ kg} \)
The orbital magnetic moment of the electron is
\( M_0 = \frac{1}{2} evr \)
\( = \frac{1}{2} (1.6 \times 10^{-19}) (2 \times 10^6) (5.3 \times 10^{-11}) \)
\( = 8.48 \times 10^{-24} \text{ A}\cdot\text{m}^2 \)
The angular momentum of the electron is
\( L_0 = m_e vr \)
\( = (9.1 \times 10^{-31}) (2 \times 10^6) (5.3 \times 10^{-11}) \)
\( = 96.46 \times 10^{-36} = 9.646 \times 10^{-35} \text{ kg}\cdot\text{m}^2/\text{s} \)

📝 Teacher's Note: This calculation helps students understand how the motion of a tiny electron creates a magnetic field, effectively acting like a microscopic current loop.

🎯 Exam Tip: Ensure you use the correct power of 10 during multiplication; it is common for students to make addition errors in exponents.

Question 7. A paramagnetic gas has \( 2.0 \times 10^{26} \text{ atoms/m}^3 \) with atomic magnetic dipole moment of \( 1.5 \times 10^{-23} \text{ A m}^2 \) 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? (\( k_B = 1.38 \times 10^{-23} \text{ JK}^{-1} \))[Answer: \( 3.0 \times 10^3 \text{ A m}^{-1} \), No]
(Hint: Find the ratio of Thermal energy of atom of a gas ( \( 3/2 \text{ k}_B\text{T} \)) and maximum potential energy of the atom (\( mB \)) and draw your conclusion)
Answer:
Data: \( \frac{N}{V} = 2.0 \times 10^{26} \text{ atoms/m}^3 \),
\( \mu = 1.5 \times 10^{-23} \text{ Am}^2 \), \( T = 27 + 273 = 300 \text{ K} \),
\( B = 3\text{T} \), \( k_B = 1.38 \times 10^{-23} \text{ J/K} \), \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \)
(a) The maximum magnetization of the material,
\( M_z = \frac{N}{V} \mu = (2.0 \times 10^{26}) (1.5 \times 10^{-23}) \)
\( = 3 \times 10^3 \text{ A/m} \)

(b) The maximum orientation energy per atom is
\( U_m = -\mu B \cos 180^{\circ} = \mu B \)
\( = (1.5 \times 10^{-23}) (3) = \frac{4.5 \times 10^{-23}}{1.6 \times 10^{-19}} \)
\( = 2.8 \times 10^{-4} \text{ eV} \)

The average thermal energy of each atom,
\( E = \frac{3}{2} k_B T \)
where \( k_B \) is the Boltzmann constant.
\( \therefore E = 1.5 (1.38 \times 10^{-23}) (300) \)
\( = 6.21 \times 10^{-21} \text{ J} = \frac{6.21 \times 10^{-21}}{1.6 \times 10^{-19}} \)
\( = 3.9 \times 10^{-2} \text{ eV} \)
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.
In simple words: Saturation happens when all atoms point the same way. Here, the heat (thermal energy) is much stronger than the magnet's pull, so the atoms keep shaking and pointing in different directions.

📝 Teacher's Note: Use the analogy of trying to line up students (dipoles) in a windy playground (thermal energy). If the wind is too strong, they won't stay in a straight line.

🎯 Exam Tip: In part (b), always explicitly compare the magnitudes of thermal energy and magnetic potential energy to justify why saturation isn't reached.

Question 8. A magnetic needle placed in uniform magnetic field has magnetic moment of \( 2 \times 10^{-2} \text{ A m}^2 \), and moment of inertia of \( 7.2 \times 10^{-7} \text{ kg m}^2 \). It performs 10 complete oscillations in 6 s. What is the magnitude of the magnetic field ?
Answer:
Data: \( M = 2 \times 10^{-2} \text{ A}\cdot\text{m}^2 \), \( I = 7.2 \times 10^{-7} \text{ kg}\cdot\text{m}^2 \),
\( T = \frac{6}{10} = 0.6 \text{ S} \)
\( T = 2\pi \sqrt{\frac{I}{MB}} \)
The magnitude of the magnetic field is
\( B = \frac{4\pi^2 I}{M T^2} \)
\( = \frac{(4)(3.14)^2 (7.2 \times 10^{-7})}{(2 \times 10^{-2})(0.6)^2} \)
\( = 3.943 \times 10^{-3} \text{ T} = 3.943 \text{ mT} \)
In simple words: We find how strong the magnetic field is by measuring how fast a magnetic needle swings back and forth, just like a pendulum.

📝 Teacher's Note: Remind students that frequency is oscillations per second, while Time Period (T) used in the formula is seconds per oscillation.

🎯 Exam Tip: Don't forget to square the Time Period (T) when rearranging the formula to solve for the magnetic field (B).

Question 9. A short bar magnet is placed in an external magnetic field of 700 gauss. 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.
Answer:
Data : \( B = 700 \text{ gauss} = 0.07 \text{ tesla} \), \( \theta = 30^{\circ} \),
\( \tau = 0.014 \text{ N}\cdot\text{m} \)
\( \tau = MB \sin \theta \)
The magnetic moment of the magnet is
\( M = \frac{\tau}{B \sin \theta} = \frac{(0.014)}{(0.07)(\sin 30^{\circ})} = 0.4 \text{ A}\cdot\text{m}^2 \)
The most stable state of the bar magnet is for \( \theta = 0^{\circ} \).
It is in the most unstable state when \( \theta = 180^{\circ} \). Thus, the work done in moving the bar magnet from 0° to 180° is
\( W = MB(\cos \theta_0 - \cos \theta) \)
\( = MB (\cos 0^{\circ} - \cos 180^{\circ}) \)
\( = MB [1 - (-1)] \)
\( = 2 MB = (2) (0.4) (0.07) \)
\( = 0.056 \text{ J} \)
This the required work done.
In simple words: It takes effort (work) to turn a magnet so it points exactly opposite to the magnetic field lines. We calculated this energy as 0.056 Joules.

📝 Teacher's Note: Explain that "stable" means the magnet is aligned with the field, while "unstable" means it is perfectly anti-aligned.

🎯 Exam Tip: Always convert gauss to tesla (\( 1 \text{ gauss} = 10^{-4} \text{ T} \)) before starting calculations to avoid magnitude errors.

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 0.1 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. (\( g = 9.8 \text{ m s}^{-2} \))
Answer:
Data: \( M = 0.2\text{g} = 2 \times 10^{-4}\text{kg} \), \( q_m = 20 \text{ A}\cdot\text{m} \), \( g = 9.8 \text{ m/s}^2 \)

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.
\( \therefore (Mg) (\frac{L}{2}) = (q_m B_v) L \)
The vertical component of the Earth’s magnetic field,
\( B_v = \frac{Mg}{2q_m} = \frac{(2 \times 10^{-4})(9.8)}{2(20)} = 4.9 \times 10^{-5} \text{ T} \)
In simple words: Earth's magnetic field pulls the needle downward. By putting a tiny weight on the other side, we balance it like a seesaw to find how strong the pull is.

📝 Teacher's Note: This problem is a great application of torque balance (\( \tau = \text{Force} \times \text{distance} \)). Ensure students see that the length \( L \) cancels out.

🎯 Exam Tip: Remember that "dip" refers to the vertical component of the Earth's field acting on the poles of the needle.

Question 11. The susceptibility of a paramagnetic material is \( \chi \) at 27° C. At what temperature its susceptibility be \( \frac{\chi}{3} \) ?
Answer:
Data: \( \chi_{m1} = \chi \), \( T_1 = 27^{\circ}\text{C} = 300 \text{ K} \), \( \chi_{m2} = \frac{\chi}{3} \)
By Curie’s law,
\( M_z = C \frac{B_0}{T} \)
Since \( M_z = \chi_m H = B_0 = \mu_0 H \)
\( \chi_m H = C \frac{\mu_0 H}{T} \)

\( \implies \chi_m = C \frac{\mu_0}{T} \)

\( \implies \chi_m \propto \frac{1}{T} \)

\( \implies \frac{\chi_{m1}}{\chi_{m2}} = \frac{T_2}{T_1} \)

\( \implies T_2 = \frac{\chi_{m1}}{\chi_{m2}} \times T_1 = 3 \frac{\chi}{\chi} \times 300 = 900 \text{ K} = 627^{\circ}\text{C} \)
This gives the required temperature.
In simple words: Curie's law says that as a material gets hotter, it becomes less magnetic. To make it three times less magnetic, we had to increase the temperature.

📝 Teacher's Note: Emphasize that Curie's law only works with absolute temperature in Kelvin. Forgetting to convert Celsius to Kelvin is the most frequent mistake here.

🎯 Exam Tip: Always convert your final answer back to Celsius if the question provides the initial temperature in Celsius.

Activity (Textbook Page No. 251)

Question 1. You have already studied in earlier classes that a short bar magnet suspended freely always aligns in North South direction. Now if you try to forcefully move and bring it in the direction along East West and leave it free, you will observe that the magnet starts turning about the axis of suspension. Do you know from where does the torque which is necessary for rotational motion come from? (as studied in rotational dynamics a torque is necessary for rotational motion).
Answer:
Suspend a short bar magnet such that it can rotate freely in a horizontal plane. Let it come to rest along the magnetic meridian. Rotate the magnet through some angle and release it. You will see that the magnet turns about the vertical axis in trying to return back to its equilibrium position along the magnetic meridian. Where does the torque for the rotational motion come from?
Take another bar magnet and bring it near the suspended magnet resting in the magnetic meridian. Observe the interaction between the like and unlike poles of the two magnets facing each other. Does the suspended magnet rotate continuously or rotate through certain angle and remain stable? Note down your observations and conclusions.
In simple words: The torque comes from the Earth's magnetic field acting on the poles of the magnet, pulling it back into alignment with the Earth's North-South poles.

📝 Teacher's Note: This activity demonstrates the concept of a restoring torque, similar to a spring being pulled and released.

🎯 Exam Tip: Mention that the torque is zero when the magnet is perfectly aligned with the magnetic meridian (\( \theta = 0 \)).

Do You Know (Textbook Page No. 255)

Effective magneton numbers for iron group ions (No. of Bohr magnetons)

Courtsey: Introduction to solid state physics by Charles Kittel, pg. 306 )
These magnetic moments are calculated from the experimental value of magnetic susceptibility. In several ions the magnetic moment is due to both orbital and spin angular momenta.
Answer:
In terms of Bohr magneton (µ_B), the effective magnetic moments of some iron group ions are as follows. In several cases, the magnetic moment is due to both orbital and spin angular momenta.
 

Remember This (Textbook Page No. 256)

Question 1. Permeability and Permittivity: Magnetic Permeability is a term analogous to permittivity in electrostatics. It basically tells us about the number of magnetic lines of force that are passing through a given substance when it is kept in an external magnetic field. The number is the indicator of the behaviour of the material in magnetic field. For superconductors \( \chi = -1 \). If you substitute in the Eq. (11.18), it is observed that permeability of material \( \mu = 0 \). This means no magnetic lines will pass through the superconductor. Magnetic Susceptibility (\( \chi \)) is the indicator of measure of the response of a given material to the external applied magnetic field. In other words it indicates as to how much magnetization will be produced in a given substance when kept in an external magnetic field. Again it is analogous to electrical susceptibility. This means when the substance is kept in a magnetic field, the atomic dipole moments either align or oppose the external magnetic field. If the atomic dipole moments of the substance are opposing the field, \( \chi \) is observed to be negative, and if the atomic dipole moments align themselves in the direction of field, \( \chi \) is observed to be positive. The number of atomic dipole moments of getting aligned in the direction of the applied magnetic field is proportional to \( \chi \). It is large for soft iron (\( \chi > 1000 \)).
Answer:
Magnetic permeability is analogous to electric permittivity, both indicating the extent to which a material permits a field to pass through or permeate into the material. For a superconductor, \( \chi = -1 \) which makes \( \mu = 0 \), so that a superconductor does not allow magnetic field lines to pass through it.
Magnetic susceptibility (\( \chi \)), analogous to electrical susceptibility, is a measure of the response of a given material to an applied magnetic field. That is, it indicates the extent of the magnetization produced in the material when it is placed in an external magnetic field. \( \chi \) is positive when the atomic dipole moments align themselves in the direction of the applied field; \( \chi \) is negative when the atomic dipole moments align antiparallel to the field. \( \chi \) is large for soft iron (\( \chi > 1000 \)).
In simple words: Permeability shows how easily magnetic lines can pass through something, while susceptibility shows how easily the material itself can become a magnet.

📝 Teacher's Note: Use the analogy of a door: high permeability means the door is wide open for magnetic lines, while a superconductor (\( \mu=0 \)) is a brick wall that blocks them completely.

🎯 Exam Tip: Remember the specific value \( \chi = -1 \) for superconductors; it's a common objective question.

Use Your Brain Power (Textbook Page No. 259)

Question 1. Classify the following atoms as diamagnetic or paramagnetic.
H, O, Zn, Fe, F, Ar, He
(Hint : Write down their electronic configurations)
Is it true that all substances with even number of electrons are diamagnetic?
Answer:

It can be seen that all substances with an even number of electrons are not necessarily diamagnetic.
In simple words: Whether an atom is magnetic depends on whether its electrons are paired up. Even if there's an even number of electrons, if some are not paired, the atom can be magnetic.

📝 Teacher's Note: Oxygen is a prime example (even electrons but paramagnetic) because of its two unpaired electrons in the 2p orbitals, which contradicts the simple "even/odd" rule.

🎯 Exam Tip: Ferromagnetism is a special case of paramagnetism with extremely high alignment; for Iron (Fe), specify it is ferromagnetic rather than just paramagnetic.

Do You Know (Textbook Page No. 260)

Question 1. Exchange Interaction: This exchange interaction in stronger than usual dipole-dipole interaction by an order of magnitude. Due to this exchange interaction, all the atomic dipole moments in a domain get aligned with each other. Find out more about the origin of exchange interaction.
Answer:
Exchange Interaction :
Quantum mechanical exchange interaction between two neighbouring spin magnetic moments in a ferromagnetic material arises as a consequence of the overlap between the magnetic orbitals of two adjacent atoms. The exchange interaction in particular for 3d metals is stronger than the dipole-dipole interaction by an order of magnitude. Due to this, all the atomic dipole moments in a domain get aligned with each other and each domain is spontaneously magnetized to saturation. (Quantum mechanics and exchange interaction are beyond the scope of the syllabus.)
In simple words: This is a powerful "group-think" for atoms. Because they are so close together, their electron spins force each other to point in the same direction, creating strongly magnetic domains.

📝 Teacher's Note: Explain that "Exchange Interaction" is a quantum effect that doesn't have a direct classical equivalent, but it's the reason why magnets work at room temperature.

🎯 Exam Tip: Focus on the result of this interaction: the spontaneous alignment of dipoles within a "domain."

Use Your Brain Power (Textbook Page No. 262)

Question 1. What does the area inside the curve B - H (hysteresis curve) indicate?
Answer:
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.
In simple words: The space inside the loop shows how much energy is wasted as heat when we magnetize and then demagnetize the material.

📝 Teacher's Note: Relate this to friction. Just as moving an object back and forth wastes energy due to friction, flipping magnetic domains wastes energy due to magnetic "friction."

🎯 Exam Tip: Remember the units: Area = Energy loss per unit volume (\( \text{J/m}^3 \)).

Do You Know (Textbook Page No. 262)

Question 1. What is soft magnetic material?
Soft ferromagnetic materials can be easily magnetized and demagnetized.
Hysteresis loop for hard and soft ferramagnetic materials.
Answer:
A soft magnetic material, usually iron-based, has high permeability, low retentivity and low coercivity. In other words, it does not have appreciable hysteresis, i.e., its hysteresis loop is very narrow. Such a material magnetizes and demagnetizes more easily, by small external fields.
In simple words: Soft magnetic materials are "easy come, easy go." They become magnetic very quickly but lose that magnetism just as fast once the external magnet is removed.

📝 Teacher's Note: Soft magnetic materials like soft iron are used in electromagnets and transformers because we want their magnetism to change instantly with the current.

🎯 Exam Tip: Key features to mention for soft materials: high permeability, low retentivity, and low coercivity.

Do You Know (Textbook Page No. 263)

Question 1. There are different types of shielding available like electrical and acoustic shielding apart from magnetic shielding discussed above. An electrical insulator functions as an electrical barrier or shield and comes in a wide array of materials. Normally the electrical wires used in our households are also shielded. In the case of audio recording, it is necessary to reduce other stray sounds which may interfere with the sound to be recorded. So the recording studios are sound insulated using acoustic material.
Answer:
There are different types of shielding, such as electrical, electromagnetic, magnetic, RF (radio frequency), and acoustic, to shield a given space or sensitive instrument from unwanted fields of each type.
In simple words: Just like wearing a coat protects you from the cold, shielding protects sensitive equipment from unwanted electricity, noise, or magnetism.

📝 Teacher's Note: Ask students where they have seen shielding in real life—for example, the lead apron at a dentist's office or the rubber coating on power cables.

🎯 Exam Tip: Understand that magnetic shielding works by redirecting the magnetic field lines through the shielding material rather than blocking them like a wall.