Contents
Class 12 Physics
Chapter 5: Magnetism & Matter
Targeting 70/70? Master the theoretical core of magnetism with these notes, tricks, and important questions.
🚀 Student Mindset Check:
Unlike "Moving Charges," this chapter is less about complex calculus and more about conceptual understanding and analogies. If you mastered Electrostatics (Chapter 1), this chapter is 80% copy-paste! We will use that to your advantage.
📚 What We Will Cover
- The Bar Magnet & Field Lines
- The "Electrostatics Analogy" (Super Trick)
- Torque & Potential Energy
- Magnetic Properties (Dia, Para, Ferro)
- Temperature Dependence (Curie's Law)
- Important Exam Questions
1. The Bar Magnet & Equivalent Solenoid
A bar magnet is a dipole with a North Pole and a South Pole. The most important property is that monopoles do not exist (you cannot isolate North from South).
Properties of Magnetic Field Lines
This is a favorite 2-mark question in board exams. Remember these key points:
- Continuous Loops: Unlike electric field lines, magnetic lines form closed continuous loops. They go from N to S outside the magnet, and S to N inside.
- Tangent Rule: The tangent to the field line at any point gives the direction of the net magnetic field ($\vec{B}$) at that point.
- No Intersection: Two lines never cross. If they did, there would be two directions of the magnetic field at one point, which is impossible.
- Density: Crowded lines = Strong Field. Spaced out lines = Weak Field.
Bar Magnet as an Equivalent Solenoid
(Qualitative Treatment)
Why do we say a solenoid acts like a bar magnet? Because when current flows through a solenoid, the magnetic field lines produced look exactly identical to those of a bar magnet.
You can verify this by cutting a solenoid into thin slices. Each slice acts as a current loop (a magnetic dipole). Summing them up gives a net magnetic moment $\vec{M}$ along the axis, just like a bar magnet.
🔥 THE GOLDEN TRICK: Electrostatics Analogy
Don't memorize new formulas! Just translate Chapter 1 formulas.
| Electrostatics (Ch 1) | Magnetism (Ch 5) |
|---|---|
| Constant: $1/4\pi\epsilon_0$ | Constant: $\mu_0/4\pi$ |
| Charge: $q$ | Pole Strength: $q_m$ (hypothetical) |
| Dipole Moment: $\vec{p}$ | Magnetic Moment: $\vec{M}$ |
| Field: $\vec{E}$ | Field: $\vec{B}$ |
2. Magnetic Field Intensity & Torque
Based on the analogy above, we can directly write the formulas required for numericals.
A. Field on Axial Line
For a short bar magnet (where length $2l \ll r$):
$$ B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} $$Direction: Along $\vec{M}$ (South to North).
B. Field on Equatorial Line
For a short bar magnet:
$$ B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3} $$Direction: Opposite to $\vec{M}$ (North to South).
Memorize: $B_{axial} = 2 \times B_{equatorial}$ (for the same distance $r$).
C. Torque on a Magnetic Dipole
When a bar magnet is placed in a uniform magnetic field $\vec{B}$ at an angle $\theta$, it experiences a torque.
$$ \vec{\tau} = \vec{M} \times \vec{B} $$ $$ \tau = MB \sin\theta $$- Max Torque: At $\theta = 90^\circ$ ($\tau = MB$).
- Min Torque (Stable Equilibrium): At $\theta = 0^\circ$ ($\tau = 0$).
- Min Torque (Unstable Equilibrium): At $\theta = 180^\circ$ ($\tau = 0$).
D. Potential Energy
Work done in rotating the dipole is stored as potential energy:
$$ U = -\vec{M} \cdot \vec{B} = -MB \cos\theta $$3. Magnetic Properties of Materials
This is the theoretical heavyweight of the chapter. CBSE loves asking differences between these materials.
Key Definitions
- Magnetization ($I$ or $M$): Net magnetic moment per unit volume.
- Magnetic Intensity ($H$): The capability of the magnetizing field ($H = nI_{current}$).
- Susceptibility ($\chi$): How easily a substance can be magnetized ($\chi = I/H$).
- Permeability ($\mu$): Ability to allow field lines to pass through.
Relation: $\mu_r = 1 + \chi$
Comparison: Dia, Para, and Ferro Magnetic
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Behavior in non-uniform field | Moves from Strong $\to$ Weak field (Repelled) | Moves from Weak $\to$ Strong field (Weakly Attracted) | Moves from Weak $\to$ Strong field (Strongly Attracted) |
| Susceptibility ($\chi$) | Small, Negative ($-1 \le \chi < 0$) | Small, Positive ($\epsilon$) | Very Large, Positive ($\approx 1000$) |
| Relative Permeability ($\mu_r$) | Less than 1 ($0 < \mu_r < 1$) | Slightly > 1 | Very Large ($\gg 1$) |
| Effect of Temperature | Independent of Temp | Inversely proportional ($\chi \propto 1/T$) | Complex (decreases with temp) |
| Examples | Bismuth, Copper, Water, Gold | Aluminum, Sodium, Oxygen (at STP) | Iron, Nickel, Cobalt, Gadolinium |
🧠 Memory Mnemonics
1. DIA = DIE (Go away). It repels field lines. It hates the magnet.
2. PARA = PARALLEL. It aligns parallel to the field but weakly.
3. FERRO = FEROCIOUS. It aligns strongly and aggressively.
For Susceptibility signs: Dia is Negative (Pessimistic), Para/Ferro are Positive (Optimistic).
1. DIA = DIE (Go away). It repels field lines. It hates the magnet.
2. PARA = PARALLEL. It aligns parallel to the field but weakly.
3. FERRO = FEROCIOUS. It aligns strongly and aggressively.
For Susceptibility signs: Dia is Negative (Pessimistic), Para/Ferro are Positive (Optimistic).
4. Effect of Temperature
Curie's Law (Paramagnetism)
For paramagnetic substances, magnetization is inversely proportional to absolute temperature.
$$ \chi = C \frac{\mu_0}{T} $$Where $C$ is Curie's constant. As distinct dipoles get hotter, thermal agitation messes up their alignment, reducing magnetism.
Curie Temperature ($T_c$) for Ferromagnets
Ferromagnetism depends on "Domains" (groups of aligned atoms). When you heat a ferromagnet, these domains break down. At a specific temperature called Curie Temperature ($T_c$), a ferromagnetic substance becomes Paramagnetic.
Example: For Iron, $T_c \approx 1043 K$. Above this, iron is no longer a strong magnet.
5. Important Q&A for Exam
Q1: Why do magnetic field lines never intersect?
Answer: If two field lines intersected at a point, there would be two tangents at that point. This implies two different directions of the net magnetic field at the same point, which is physically impossible.
Q2: A short bar magnet placed with its axis at $30^\circ$ with a uniform magnetic field of $0.25 T$ experiences a torque of magnitude $4.5 \times 10^{-2} J$. What is the magnitude of the magnetic moment of the magnet?
Answer:
Given: $\theta = 30^\circ$, $B = 0.25 T$, $\tau = 4.5 \times 10^{-2} Nm$.
Formula: $\tau = MB \sin\theta$
$M = \frac{\tau}{B \sin\theta}$
$M = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5}$
$M = 0.36 \, A m^2$.
Given: $\theta = 30^\circ$, $B = 0.25 T$, $\tau = 4.5 \times 10^{-2} Nm$.
Formula: $\tau = MB \sin\theta$
$M = \frac{\tau}{B \sin\theta}$
$M = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5}$
$M = 0.36 \, A m^2$.
Q3: How does the magnetic susceptibility of a diamagnetic substance depend on temperature?
Answer: It is practically independent of temperature. The induced dipole moment in diamagnetism arises from the change in orbital motion of electrons, which is not significantly affected by thermal agitation.
Q4: Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?
Answer: Paramagnetism comes from the alignment of permanent atomic dipoles. At lower temperatures, the thermal randomizing effect (thermal agitation) is reduced, allowing the dipoles to align more effectively with the external field, thus increasing magnetization.
6. Deep Dive: Magnetization & Magnetic Intensity
To score full marks in 3-mark or 5-mark conceptual questions, you need to understand the precise difference between $\vec{M}$, $\vec{B}$, and $\vec{H}$. Many students confuse these.
The Logic Chain
- Solenoid Field ($\vec{B}_0$): When we pass current through a solenoid, we generate a vacuum field: $\vec{B}_0 = \mu_0 n I$.
- Material Insertion: We place a material inside. The material gets magnetized.
- Induced Field ($\vec{B}_m$): The material creates its own field due to dipole alignment: $\vec{B}_m = \mu_0 \vec{M}$.
- Total Field ($\vec{B}$): The net field is the sum: $\vec{B} = \vec{B}_0 + \vec{B}_m$.
Combining these, we get the master equation:
$$ \vec{B} = \mu_0 (\vec{H} + \vec{M}) $$Here, $\vec{H}$ (Magnetic Intensity) represents the external agency causing the magnetization, while $\vec{M}$ represents the material's response.
Hard vs. Soft Ferromagnets
Ferromagnetic materials are further classified based on how "stubborn" they are. This is crucial for application-based questions.
| Feature | Soft Ferromagnets | Hard Ferromagnets |
|---|---|---|
| Nature | Easily magnetized and easily demagnetized. | Difficult to magnetize, but once magnetized, they stay that way. |
| Retentivity | Low (Loses magnetism when current stops). | High (Retains magnetism). |
| Coercivity | Low (Easy to reverse). | High (Hard to reverse). |
| Uses | Electromagnets, Transformers, Telephone diaphragms. | Permanent Magnets, Loudspeakers, Electric meters. |
| Material Examples | Soft Iron, Permalloy. | Alnico (Al-Ni-Co), Steel, Ticonal. |
💡 Trick to Remember:
Soft Iron is "soft-hearted"—it lets go of its magnetism easily (good for temporary electromagnets).
Steel is "hard-headed"—it is stubborn and keeps its magnetism (good for permanent magnets).
7. Gauss's Law in Magnetism
In electrostatics, we know $\oint \vec{E} \cdot d\vec{s} = q/\epsilon_0$. However, magnetism is different.
$$ \oint \vec{B} \cdot d\vec{s} = 0 $$Physical Significance (Exam Favorite):
- This equation implies that isolated magnetic poles (monopoles) do not exist.
- Any magnetic field line that enters a closed surface must also leave it.
- The net magnetic flux through any closed surface is always zero.
8. Advanced Numerical Practice (Level 2)
If you want to secure the top rank, solve these application-based problems.
Q5: A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per meter, calculate (a) H, (b) M, (c) B.
Solution:
Given: $n = 1000 \, m^{-1}$, $I = 2A$, $\mu_r = 400$.
(a) Magnetic Intensity ($H$):
$H = nI = 1000 \times 2 = 2000 \, A/m$.
(b) Magnetization ($M$):
Since $\mu_r = 1 + \chi$, we have $\chi = 399$.
$M = \chi H = 399 \times 2000 \approx 8 \times 10^5 \, A/m$.
(Approximation trick: Since $\mu_r \gg 1$, $M \approx H(\mu_r) \approx H \mu_r$ is rarely used, stick to $\chi$). Actually, for high $\mu_r$, $M \approx H \mu_r$ is roughly accepted but $M = (\mu_r - 1)H$ is accurate.
(c) Magnetic Field ($B$):
$B = \mu_r \mu_0 H$
$B = 400 \times (4\pi \times 10^{-7}) \times 2000$
$B = 8 \times 10^5 \times 4\pi \times 10^{-7} = 32\pi \times 10^{-2} \approx 1.0 \, T$.
Given: $n = 1000 \, m^{-1}$, $I = 2A$, $\mu_r = 400$.
(a) Magnetic Intensity ($H$):
$H = nI = 1000 \times 2 = 2000 \, A/m$.
(b) Magnetization ($M$):
Since $\mu_r = 1 + \chi$, we have $\chi = 399$.
$M = \chi H = 399 \times 2000 \approx 8 \times 10^5 \, A/m$.
(Approximation trick: Since $\mu_r \gg 1$, $M \approx H(\mu_r) \approx H \mu_r$ is rarely used, stick to $\chi$). Actually, for high $\mu_r$, $M \approx H \mu_r$ is roughly accepted but $M = (\mu_r - 1)H$ is accurate.
(c) Magnetic Field ($B$):
$B = \mu_r \mu_0 H$
$B = 400 \times (4\pi \times 10^{-7}) \times 2000$
$B = 8 \times 10^5 \times 4\pi \times 10^{-7} = 32\pi \times 10^{-2} \approx 1.0 \, T$.
Q6: A domain in ferromagnetic iron is in the form of a cube of side length $1 \mu m$. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetization of the domain. The atomic mass of iron is 55 g/mole and its density is $7.9 \, g/cm^3$. Assume each iron atom has a dipole moment of $9.27 \times 10^{-24} \, Am^2$.
Solution (Step-by-Step):
1. Volume of Domain ($V$):
$V = (10^{-6} m)^3 = 10^{-18} m^3 = 10^{-12} cm^3$.
2. Mass of Domain ($m$):
$m = \text{Volume} \times \text{Density} = 10^{-12} \times 7.9 = 7.9 \times 10^{-12} g$.
3. Number of Atoms ($N$):
$N = \frac{\text{Mass}}{\text{Atomic Mass}} \times N_A$
$N = \frac{7.9 \times 10^{-12}}{55} \times 6.023 \times 10^{23} \approx 8.65 \times 10^{10} \text{ atoms}$.
4. Max Dipole Moment ($M_{tot}$):
$M_{tot} = N \times m_{atom} = 8.65 \times 10^{10} \times 9.27 \times 10^{-24}$
$M_{tot} \approx 8.0 \times 10^{-13} \, Am^2$.
5. Magnetization ($M$):
$M = \frac{M_{tot}}{Volume} = \frac{8.0 \times 10^{-13}}{10^{-18}} = 8.0 \times 10^5 \, Am^{-1}$.
1. Volume of Domain ($V$):
$V = (10^{-6} m)^3 = 10^{-18} m^3 = 10^{-12} cm^3$.
2. Mass of Domain ($m$):
$m = \text{Volume} \times \text{Density} = 10^{-12} \times 7.9 = 7.9 \times 10^{-12} g$.
3. Number of Atoms ($N$):
$N = \frac{\text{Mass}}{\text{Atomic Mass}} \times N_A$
$N = \frac{7.9 \times 10^{-12}}{55} \times 6.023 \times 10^{23} \approx 8.65 \times 10^{10} \text{ atoms}$.
4. Max Dipole Moment ($M_{tot}$):
$M_{tot} = N \times m_{atom} = 8.65 \times 10^{10} \times 9.27 \times 10^{-24}$
$M_{tot} \approx 8.0 \times 10^{-13} \, Am^2$.
5. Magnetization ($M$):
$M = \frac{M_{tot}}{Volume} = \frac{8.0 \times 10^{-13}}{10^{-18}} = 8.0 \times 10^5 \, Am^{-1}$.
Q7: What happens if a bar magnet is cut into two pieces: (a) Transverse to its length, (b) Along its length?
Answer:
(a) Transverse Cut (Cutting length in half):
Pole strength ($m$) remains same. Length becomes $L/2$.
New Magnetic Moment $M' = m \times (L/2) = M/2$.
(b) Longitudinal Cut (Cutting along the axis):
Pole strength becomes $m/2$. Length remains $L$.
New Magnetic Moment $M' = (m/2) \times L = M/2$.
Conclusion: In both cases, the magnetic moment is halved!
(a) Transverse Cut (Cutting length in half):
Pole strength ($m$) remains same. Length becomes $L/2$.
New Magnetic Moment $M' = m \times (L/2) = M/2$.
(b) Longitudinal Cut (Cutting along the axis):
Pole strength becomes $m/2$. Length remains $L$.
New Magnetic Moment $M' = (m/2) \times L = M/2$.
Conclusion: In both cases, the magnetic moment is halved!
🎉 Final Study Tip
Before the exam, practice drawing the field lines of a Bar Magnet and a Solenoid side-by-side. Make sure you can write the definition of "Domains" for ferromagnetism. These are the high-probability subjective questions!
🎯 The "Exam Dominator" Question Bank (40+ Qs)
Active Recall Method: Try to answer before clicking to reveal!
🔥 Part 1: Rapid Fire (1 Mark) – Speed & Accuracy
Q1. What is the SI unit of Magnetic Moment ($\vec{M}$)?
$Am^2$ (Ampere-meter squared) or $J/T$ (Joule per Tesla).
Q2. Is magnetic moment a scalar or vector quantity?
Vector. Direction is from South Pole to North Pole inside the magnet.
Q3. What is the value of net magnetic flux passing through a closed surface?
Zero. (Gauss's Law for Magnetism: $\oint \vec{B} \cdot d\vec{s} = 0$).
Q4. What is the unit of Magnetic Field Intensity ($\vec{H}$)?
Bigyanbook Trick: H and M have the same unit!
Answer: $A/m$ (Ampere per meter).
Answer: $A/m$ (Ampere per meter).
Q5. Define Magnetic Susceptibility ($\chi$).
It is the ratio of intensity of magnetization ($M$) induced in the material to the magnetizing force ($H$). $\chi = M/H$.
Q6. Which material has negative susceptibility?
Diamagnetic materials. (Think: "Negative" attitude, they repel).
Q7. What is the relation between Relative Permeability ($\mu_r$) and Susceptibility ($\chi$)?
$\mu_r = 1 + \chi$.
Q8. How does the magnetic susceptibility of a paramagnetic substance depend on temperature?
Inversely proportional. $\chi \propto 1/T$ (Curie’s Law).
Q9. Does the time period of a magnet in a uniform magnetic field depend on its mass?
Yes. $T = 2\pi \sqrt{I/MB}$. Here $I$ is Moment of Inertia, which depends on mass.
Q10. What is the work done in rotating a magnet by $360^\circ$ in a magnetic field?
Zero. (Conservative field).
🧠 Part 2: Logical Reasoning (2 Marks) – The "Why" Questions
Q11. Why do magnetic field lines form continuous closed loops?
Because magnetic monopoles (isolated N or S poles) do not exist. Field lines run N to S outside and S to N inside.
Q12. Two identical bar magnets are placed one above the other with like poles together. What happens to the time period of oscillation?
$M_{net} = M + M = 2M$. $I_{net} = I + I = 2I$.
$T' = 2\pi \sqrt{2I / (2M \cdot B)} = T$.
The time period remains unchanged!
$T' = 2\pi \sqrt{2I / (2M \cdot B)} = T$.
The time period remains unchanged!
Q13. Two identical bar magnets are placed in a cross ($+$) shape. What is the net magnetic moment?
They are at $90^\circ$.
$M_{net} = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Direction is at $45^\circ$ to both.
$M_{net} = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Direction is at $45^\circ$ to both.
Q14. Why is Soft Iron used to make electromagnets?
Trick: Soft Iron = "Soft" Memory.
Reason: It has low retentivity (loses magnetism quickly when current stops) and low coercivity (easy to demagnetize), plus high permeability.
Reason: It has low retentivity (loses magnetism quickly when current stops) and low coercivity (easy to demagnetize), plus high permeability.
Q15. Why is Steel used to make permanent magnets?
Because Steel has high retentivity (holds magnetism strongly) and high coercivity (hard to destroy the magnetism).
Q16. A magnetic needle free to rotate in a vertical plane orients itself vertically at a certain place on Earth. Where is this place?
At the Magnetic Poles. (The field lines are perpendicular to the surface there).
Q17. What happens to the magnetic moment if a hole is drilled through the center of a bar magnet?
The volume decreases, so the total number of atomic dipoles decreases. Thus, the Magnetic Moment ($M$) decreases slightly.
Q18. Define "Curie Temperature".
The temperature above which a ferromagnetic substance loses its ferromagnetism and becomes paramagnetic.
Q19. Can we have a magnetic field without a magnetic pole?
Yes! A current-carrying toroid or an infinite straight wire produces a magnetic field but has no poles (the lines are circular loops with no ends).
Q20. Why does a paramagnetic liquid in a U-tube rise when one limb is placed between strong magnetic poles?
Paramagnetic substances move from weaker to stronger parts of the field. The liquid is attracted to the strong field region between the poles.
🧮 Part 3: Numerical Warfare – Apply the Formulas
Q21. A short bar magnet has a magnetic moment of $0.48 J/T$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the center of the magnet on the axis.
Formula: $B = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$
Given: $M = 0.48$, $r = 0.1 m$.
$B = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3}$
$B = 10^{-7} \times \frac{0.96}{0.001} = 0.96 \times 10^{-4} T$.
Direction: Along the magnet's N-S axis.
Given: $M = 0.48$, $r = 0.1 m$.
$B = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3}$
$B = 10^{-7} \times \frac{0.96}{0.001} = 0.96 \times 10^{-4} T$.
Direction: Along the magnet's N-S axis.
Q22. In the previous question, what if the point is on the equatorial lines?
Concept: $B_{equatorial} = \frac{1}{2} B_{axial}$.
$B = \frac{0.96 \times 10^{-4}}{2} = 0.48 \times 10^{-4} T$.
Direction: Opposite to the magnetic moment (N to S).
$B = \frac{0.96 \times 10^{-4}}{2} = 0.48 \times 10^{-4} T$.
Direction: Opposite to the magnetic moment (N to S).
Q23. A magnet of moment $M$ is bent into a semi-circular arc. What is the new magnetic moment?
Classic Trick!
Original Length $L = \pi R \Rightarrow R = L/\pi$.
Original $M = m \times L$.
New Distance between poles = Diameter = $2R = 2L/\pi$.
New $M' = m \times (2L/\pi) = \frac{2}{\pi} (mL) = \frac{2M}{\pi}$.
Original Length $L = \pi R \Rightarrow R = L/\pi$.
Original $M = m \times L$.
New Distance between poles = Diameter = $2R = 2L/\pi$.
New $M' = m \times (2L/\pi) = \frac{2}{\pi} (mL) = \frac{2M}{\pi}$.
Q24. A bar magnet of magnetic moment $1.5 J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 T$. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?
Initial angle $\theta_1 = 0^\circ$. Final angle $\theta_2 = 90^\circ$.
$W = -MB (\cos \theta_2 - \cos \theta_1)$
$W = -MB (\cos 90 - \cos 0) = -MB(0 - 1) = MB$.
$W = 1.5 \times 0.22 = 0.33 J$.
$W = -MB (\cos \theta_2 - \cos \theta_1)$
$W = -MB (\cos 90 - \cos 0) = -MB(0 - 1) = MB$.
$W = 1.5 \times 0.22 = 0.33 J$.
Q25. In the question above, what is the torque in the final position?
$\tau = MB \sin 90^\circ = MB$.
$\tau = 1.5 \times 0.22 = 0.33 Nm$.
$\tau = 1.5 \times 0.22 = 0.33 Nm$.
⚖️ Part 4: Material Identification (Rapid Check)
Identify the material type (Dia, Para, Ferro) for Q26-Q30:
Q26. $\chi = -0.00015$
Diamagnetic (Small negative).
Q27. $\mu_r = 1.0002$
Paramagnetic (Small positive).
Q28. $\mu_r = 2500$
Ferromagnetic (Large positive).
Q29. Material is repelled by a strong magnet.
Diamagnetic.
Q30. $\chi$ decreases rapidly as Temperature increases.
Ferromagnetic (or Paramagnetic, but rapid decrease implies Ferro transition). Specifically, Paramagnetic follows $1/T$.
🧐 Part 5: Tricky Concepts & Mental Hacks
Q31. Can two magnetic field lines intersect? (Yes/No)
NO. Never.
Q32. True or False: Diamagnetism is present in all materials.
TRUE! (Shocking, right?).
Explanation: Diamagnetism arises from the intrinsic orbital motion of electrons, which all atoms have. However, in Para and Ferro materials, the stronger effects mask the weak diamagnetism.
Explanation: Diamagnetism arises from the intrinsic orbital motion of electrons, which all atoms have. However, in Para and Ferro materials, the stronger effects mask the weak diamagnetism.
Q33. What is the potential energy of a dipole at unstable equilibrium?
Unstable equilibrium is at $\theta = 180^\circ$.
$U = -MB \cos(180) = -MB(-1) = +MB$ (Maximum Potential Energy).
$U = -MB \cos(180) = -MB(-1) = +MB$ (Maximum Potential Energy).
Q34. If a solenoid is free to turn, what direction will it point?
It acts like a bar magnet. It will point North-South.
Q35. What is the value of $\chi$ for a superconductor?
Exam Favorite!
Superconductors are perfect diamagnets.
So $B = 0$ inside. $B = \mu_0(H+M) \Rightarrow H = -M \Rightarrow \chi = -1$.
Superconductors are perfect diamagnets.
So $B = 0$ inside. $B = \mu_0(H+M) \Rightarrow H = -M \Rightarrow \chi = -1$.
Q36. Does magnetic shielding exist?
Yes. By using a soft iron shell (high permeability), magnetic field lines prefer to pass through the iron shell rather than the empty space inside, shielding the interior.
Q37. Which is stronger: Electric force or Magnetic force?
Generally, Electric force is much stronger. Magnetic force is a relativistic effect of electricity.
Q38. Why do field lines crowd near the poles?
The density of field lines represents field strength. The field is strongest near the poles, so the lines are crowded.
Q39. Can a vibration magnetometer be used to compare magnetic moments?
Yes. Using $T = 2\pi \sqrt{I/MB}$. If $B$ and $I$ are known/constant, we can compare $M_1$ and $M_2$.
Q40. Final Check: Does a stationary charge experience a force in a magnetic field?
No. $F = qvB \sin\theta$. If $v=0$, Force is zero.
🌟 Student Affirmation
If you could answer 25+ of these correctly, you are in the Safe Zone.
If you answered 35+, you are in the Topper Zone.
If you answered less than 20, don't panic! Just re-read the "Distinction Table" and the Formula list in the notes above. You've got this!