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Magnetism and Matter NCERT Solutions, Complete PDF Content and Important Questions
In CBSE Class 12 Physics Chapter 5, Magnetism and Matter, you will study the fascinating magnetic properties of materials and the behavior of bar magnets. This chapter is highly scoring and critical for securing top marks in your 2026 board exams. It also builds foundational concepts frequently tested in competitive examinations like JEE and NEET. Let's simplify these core topics together!
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Chapter NameMagnetism and Matter
SubjectPhysics
Class12
BoardCBSE / State Boards
DifficultyEasy to Moderate
Exam WeightageAround 5-6 Marks (Clubbed with Chapter 4)
Learning Objectives
After completing this chapter, students will be able to:
- Explain the properties of a bar magnet and compare it with a current-carrying solenoid.
- Plot and interpret magnetic field lines for different configurations.
- Understand Gauss's Law in Magnetism and why isolated magnetic monopoles do not exist.
- Calculate the torque, work done, and potential energy of a magnetic dipole in a uniform magnetic field.
- Classify materials into diamagnetic, paramagnetic, and ferromagnetic based on their magnetic properties and susceptibility.
Key Concepts, Definitions and Formulas
- Magnetic Dipole Moment ($M$): For a bar magnet, it is the product of pole strength ($m$) and magnetic length ($2l$).
$$\mathbf{M} = m \times 2l$$
Its SI unit is $\text{A m}^2$ or $\text{J/T}$. It is a vector quantity directed from the South Pole to the North Pole. - Magnetic Field of a Short Bar Magnet:
- On its Axial Point (at distance $r$): $$B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$$
- On its Equatorial Point (at distance $r$): $$B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$$
- Torque ($\tau$) on a Magnetic Dipole: When placed in a uniform magnetic field $B$ at an angle $\theta$:
$$\tau = \mathbf{M} \times \mathbf{B} = MB \sin\theta$$ - Potential Energy ($U$) of a Magnetic Dipole:
$$U = -\mathbf{M} \cdot \mathbf{B} = -MB \cos\theta$$
Stable Equilibrium: $\theta = 0^\circ \implies U = -MB$ (Minimum energy)
Unstable Equilibrium: $\theta = 180^\circ \implies U = +MB$ (Maximum energy) - Gauss's Law for Magnetism: The net magnetic flux through any closed surface is always zero.
$$\oint \mathbf{B} \cdot d\mathbf{A} = 0$$ - Magnetic Intensity ($H$) and Intensity of Magnetization ($M_m$):
$$B = \mu_0(H + M_m)$$ - Magnetic Susceptibility ($\chi_m$): Measures how easily a substance gets magnetized.
$$\chi_m = \frac{M_m}{H}$$ - Relative Permeability ($\mu_r$): Related to susceptibility by the formula:
$$\mu_r = 1 + \chi_m$$
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Full NCERT Solutions
(Strictly based on the latest rationalized syllabus for CBSE Board Exam 2026)
Question 1: A short bar magnet placed with its axis at $30^\circ$ with a uniform external magnetic field of $0.25 \text{ T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \text{ J}$. What is the magnitude of magnetic moment of the magnet?
Answer:
Step 1: Identify given values.
Angle, $\theta = 30^\circ$
External magnetic field, $B = 0.25 \text{ T}$
Torque, $\tau = 4.5 \times 10^{-2} \text{ J}$
Step 2: Apply the torque formula.
$$\tau = MB \sin\theta$$
Step 3: Rearrange to find the magnetic moment ($M$) and calculate.
$$M = \frac{\tau}{B \sin\theta}$$
$$M = \frac{4.5 \times 10^{-2}}{0.25 \times \sin(30^\circ)}$$
$$M = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} = 0.36 \text{ A m}^2$$
Final Answer: The magnitude of the magnetic moment is $0.36 \text{ A m}^2$ (or $\text{J/T}$).
Step 1: Identify given values.
Angle, $\theta = 30^\circ$
External magnetic field, $B = 0.25 \text{ T}$
Torque, $\tau = 4.5 \times 10^{-2} \text{ J}$
Step 2: Apply the torque formula.
$$\tau = MB \sin\theta$$
Step 3: Rearrange to find the magnetic moment ($M$) and calculate.
$$M = \frac{\tau}{B \sin\theta}$$
$$M = \frac{4.5 \times 10^{-2}}{0.25 \times \sin(30^\circ)}$$
$$M = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} = 0.36 \text{ A m}^2$$
Final Answer: The magnitude of the magnetic moment is $0.36 \text{ A m}^2$ (or $\text{J/T}$).
Question 2: A short bar magnet of magnetic moment $M = 0.32 \text{ J/T}$ is placed in a uniform magnetic field of $0.15 \text{ T}$. If the magnet is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Answer:
Step 1: Identify given values.
Magnetic moment, $M = 0.32 \text{ J/T}$
Magnetic field, $B = 0.15 \text{ T}$
Step 2: Calculate for Stable Equilibrium.
(a) Stable equilibrium occurs when the magnetic moment is parallel to the magnetic field, meaning $\theta = 0^\circ$.
Potential energy formula:
$$U = -MB \cos\theta$$
$$U_{\text{stable}} = -0.32 \times 0.15 \times \cos(0^\circ)$$
$$U_{\text{stable}} = -0.048 \times 1 = -0.048 \text{ J}$$
Step 3: Calculate for Unstable Equilibrium.
(b) Unstable equilibrium occurs when the magnetic moment is antiparallel to the magnetic field, meaning $\theta = 180^\circ$.
$$U_{\text{unstable}} = -0.32 \times 0.15 \times \cos(180^\circ)$$
$$U_{\text{unstable}} = -0.048 \times (-1) = +0.048 \text{ J}$$
Final Answer:
* Stable orientation: aligned with the field ($\theta = 0^\circ$), Potential Energy = $-0.048 \text{ J}$
* Unstable orientation: opposite to the field ($\theta = 180^\circ$), Potential Energy = $+0.048 \text{ J}$.
Step 1: Identify given values.
Magnetic moment, $M = 0.32 \text{ J/T}$
Magnetic field, $B = 0.15 \text{ T}$
Step 2: Calculate for Stable Equilibrium.
(a) Stable equilibrium occurs when the magnetic moment is parallel to the magnetic field, meaning $\theta = 0^\circ$.
Potential energy formula:
$$U = -MB \cos\theta$$
$$U_{\text{stable}} = -0.32 \times 0.15 \times \cos(0^\circ)$$
$$U_{\text{stable}} = -0.048 \times 1 = -0.048 \text{ J}$$
Step 3: Calculate for Unstable Equilibrium.
(b) Unstable equilibrium occurs when the magnetic moment is antiparallel to the magnetic field, meaning $\theta = 180^\circ$.
$$U_{\text{unstable}} = -0.32 \times 0.15 \times \cos(180^\circ)$$
$$U_{\text{unstable}} = -0.048 \times (-1) = +0.048 \text{ J}$$
Final Answer:
* Stable orientation: aligned with the field ($\theta = 0^\circ$), Potential Energy = $-0.048 \text{ J}$
* Unstable orientation: opposite to the field ($\theta = 180^\circ$), Potential Energy = $+0.048 \text{ J}$.
Question 3: A closely wound solenoid of $800 \text{ turns}$ and area of cross-section $2.5 \times 10^{-4} \text{ m}^2$ carries a current of $3.0 \text{ A}$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Step 1: Understand the conceptual part.
A current-carrying solenoid produces a magnetic field along its axis. One end behaves like a North pole where field lines emerge, and the other behaves like a South pole where lines enter. Thus, it acts exactly like a bar magnet.
Step 2: Write down numerical values.
Number of turns, $N = 800$
Area of cross-section, $A = 2.5 \times 10^{-4} \text{ m}^2$
Current, $I = 3.0 \text{ A}$
Step 3: Calculate the magnetic dipole moment ($M$).
$$M = NIA$$
$$M = 800 \times 3.0 \times 2.5 \times 10^{-4}$$
$$M = 2400 \times 2.5 \times 10^{-4} = 0.6 \text{ J/T}$$
Final Answer: The solenoid acts like a bar magnet along its axis, and its associated magnetic moment is $0.6 \text{ J/T}$ along the direction of the axis.
Step 1: Understand the conceptual part.
A current-carrying solenoid produces a magnetic field along its axis. One end behaves like a North pole where field lines emerge, and the other behaves like a South pole where lines enter. Thus, it acts exactly like a bar magnet.
Step 2: Write down numerical values.
Number of turns, $N = 800$
Area of cross-section, $A = 2.5 \times 10^{-4} \text{ m}^2$
Current, $I = 3.0 \text{ A}$
Step 3: Calculate the magnetic dipole moment ($M$).
$$M = NIA$$
$$M = 800 \times 3.0 \times 2.5 \times 10^{-4}$$
$$M = 2400 \times 2.5 \times 10^{-4} = 0.6 \text{ J/T}$$
Final Answer: The solenoid acts like a bar magnet along its axis, and its associated magnetic moment is $0.6 \text{ J/T}$ along the direction of the axis.
Question 4: If the solenoid in Question 3 is free to turn about a vertical direction and a uniform horizontal magnetic field of $0.25 \text{ T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^\circ$ with the direction of the field?
Answer:
Step 1: Bring values from the previous question and list the new ones.
Magnetic moment from Q3, $M = 0.6 \text{ J/T}$
Horizontal magnetic field, $B = 0.25 \text{ T}$
Angle, $\theta = 30^\circ$
Step 2: Apply the torque formula.
$$\tau = MB \sin\theta$$
$$\tau = 0.6 \times 0.25 \times \sin(30^\circ)$$
$$\tau = 0.15 \times 0.5 = 0.075 \text{ N m}$$
Final Answer: The magnitude of the torque acting on the solenoid is $0.075 \text{ N m}$ (or $\text{J}$).
Step 1: Bring values from the previous question and list the new ones.
Magnetic moment from Q3, $M = 0.6 \text{ J/T}$
Horizontal magnetic field, $B = 0.25 \text{ T}$
Angle, $\theta = 30^\circ$
Step 2: Apply the torque formula.
$$\tau = MB \sin\theta$$
$$\tau = 0.6 \times 0.25 \times \sin(30^\circ)$$
$$\tau = 0.15 \times 0.5 = 0.075 \text{ N m}$$
Final Answer: The magnitude of the torque acting on the solenoid is $0.075 \text{ N m}$ (or $\text{J}$).
Question 5: A bar magnet of magnetic moment $1.5 \text{ J/T}$ lies aligned with the direction of a uniform magnetic field of $0.22 \text{ T}$. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in case (ii)?
Answer:
Given: $M = 1.5 \text{ J/T}$, $B = 0.22 \text{ T}$. Initially, the magnet is aligned with the field, so $\theta_1 = 0^\circ$.
(a) Work Done calculation:
The work done by an external agent is given by:
$$W = MB(\cos\theta_1 - \cos\theta_2)$$
* Case (i): Normal to field direction ($\theta_2 = 90^\circ$):
$$W = 1.5 \times 0.22 \times (\cos 0^\circ - \cos 90^\circ)$$
$$W = 0.33 \times (1 - 0) = 0.33 \text{ J}$$
* Case (ii): Opposite to field direction ($\theta_2 = 180^\circ$):
$$W = 1.5 \times 0.22 \times (\cos 0^\circ - \cos 180^\circ)$$
$$W = 0.33 \times (1 - (-1)) = 0.33 \times 2 = 0.66 \text{ J}$$
(b) Torque in case (ii):
When the magnet is opposite to the field direction ($\theta = 180^\circ$);
$$\tau = MB \sin(180^\circ) = 1.5 \times 0.22 \times 0 = 0$$
Final Answer:
* (a) Work required: (i) $0.33 \text{ J}$, (ii) $0.66 \text{ J}$.
* (b) The torque on the magnet in case (ii) is $0$.
Given: $M = 1.5 \text{ J/T}$, $B = 0.22 \text{ T}$. Initially, the magnet is aligned with the field, so $\theta_1 = 0^\circ$.
(a) Work Done calculation:
The work done by an external agent is given by:
$$W = MB(\cos\theta_1 - \cos\theta_2)$$
* Case (i): Normal to field direction ($\theta_2 = 90^\circ$):
$$W = 1.5 \times 0.22 \times (\cos 0^\circ - \cos 90^\circ)$$
$$W = 0.33 \times (1 - 0) = 0.33 \text{ J}$$
* Case (ii): Opposite to field direction ($\theta_2 = 180^\circ$):
$$W = 1.5 \times 0.22 \times (\cos 0^\circ - \cos 180^\circ)$$
$$W = 0.33 \times (1 - (-1)) = 0.33 \times 2 = 0.66 \text{ J}$$
(b) Torque in case (ii):
When the magnet is opposite to the field direction ($\theta = 180^\circ$);
$$\tau = MB \sin(180^\circ) = 1.5 \times 0.22 \times 0 = 0$$
Final Answer:
* (a) Work required: (i) $0.33 \text{ J}$, (ii) $0.66 \text{ J}$.
* (b) The torque on the magnet in case (ii) is $0$.
Question 6: A closely wound solenoid of $2000 \text{ turns}$ and cross-sectional area $1.6 \times 10^{-4} \text{ m}^2$, carrying a current of $4.0 \text{ A}$, is suspended through its centre. (a) What is its magnetic moment? (b) If a uniform horizontal magnetic field of $7.5 \times 10^{-2} \text{ T}$ is set up at an angle of $30^\circ$ with the axis of the solenoid, find the magnitude of torque...
Answer:
(a) Find Magnetic Moment ($M$):
$N = 2000$, $A = 1.6 \times 10^{-4} \text{ m}^2$, $I = 4.0 \text{ A}$
$$M = NIA = 2000 \times 4.0 \times 1.6 \times 10^{-4}$$
$$M = 8000 \times 1.6 \times 10^{-4} = 1.28 \text{ A m}^2$$
(b) Find Torque ($\tau$):
$B = 7.5 \times 10^{-2} \text{ T}$, $\theta = 30^\circ$
$$\tau = MB \sin\theta = 1.28 \times (7.5 \times 10^{-2}) \times \sin(30^\circ)$$
$$\tau = 0.096 \times 0.5 = 0.048 \text{ N m}$$
Final Answer: (a) Magnetic moment is $1.28 \text{ A m}^2$. (b) Torque is $0.048 \text{ N m}$.
(a) Find Magnetic Moment ($M$):
$N = 2000$, $A = 1.6 \times 10^{-4} \text{ m}^2$, $I = 4.0 \text{ A}$
$$M = NIA = 2000 \times 4.0 \times 1.6 \times 10^{-4}$$
$$M = 8000 \times 1.6 \times 10^{-4} = 1.28 \text{ A m}^2$$
(b) Find Torque ($\tau$):
$B = 7.5 \times 10^{-2} \text{ T}$, $\theta = 30^\circ$
$$\tau = MB \sin\theta = 1.28 \times (7.5 \times 10^{-2}) \times \sin(30^\circ)$$
$$\tau = 0.096 \times 0.5 = 0.048 \text{ N m}$$
Final Answer: (a) Magnetic moment is $1.28 \text{ A m}^2$. (b) Torque is $0.048 \text{ N m}$.
Extra Important Questions (Board Style)
Multiple Choice Questions (MCQs)
1. Which of the following statements is true according to Gauss's Law for magnetism?
a) Magnetic monopoles can exist under extreme conditions.
b) The net magnetic flux through any closed surface is zero.
c) Total magnetic field inside a conductor is zero.
d) Flux depends on the surface area.
Answer: (b) The net magnetic flux through any closed surface is zero.
2. A diamagnetic substance is brought near the north or south pole of a bar magnet. It will experience:
a) Attraction towards both poles
b) Repulsion from both poles
c) Attraction towards the North pole only
d) Repulsion from the North pole only
Answer: (b) Repulsion from both poles. Explanation: Diamagnetic substances create a weak magnetization opposite to the external field.
3. Magnetic susceptibility ($\chi_m$) of a paramagnetic material varies with temperature $T$ as:
a) $\chi_m \propto T$
b) $\chi_m \propto \frac{1}{T}$
c) $\chi_m \propto \frac{1}{T^2}$
d) Independent of temperature
Answer: (b) $\chi_m \propto \frac{1}{T}$ (Curie's Law).
4. The relative permeability ($\mu_r$) of a diamagnetic material is:
a) Slightly greater than 1
b) Equal to 1
c) Slightly less than 1
d) Much greater than 1
Answer: (c) Slightly less than 1.
5. What is the angle between the magnetic dipole moment vector and the equatorial magnetic field of a short bar magnet?
a) $0^\circ$
b) $90^\circ$
c) $180^\circ$
d) $45^\circ$
Answer: (c) $180^\circ$. Explanation: Equatorial magnetic field direction is opposite to the magnetic dipole moment.
Short Answer Questions (2-3 Marks)
6. State three differences between magnetic properties of diamagnetic and paramagnetic substances.
Answer:
| Property | Diamagnetic | Paramagnetic |
|---|---|---|
| Susceptibility ($\chi_m$) | Negative and small ($-1 \le \chi_m < 0$) | Positive and small ($0 < \chi_m < \epsilon$) |
| Effect of Magnet | Feebly repelled by a magnet | Feebly attracted by a magnet |
| Temperature effect | Independent of temperature | Decreases with increase in temperature |
7. Why do magnetic field lines form continuous closed loops whereas electric field lines do not?
Answer: Isolated magnetic monopoles (isolated N or S poles) do not exist; magnetic poles always occur in equal and opposite pairs. Hence, lines must emerge from the North pole and enter the South pole externally, completing the loop inside the magnet. Electric field lines emerge from isolated positive charges and terminate at negative charges because isolated electric charges can exist.
8. Define the terms (i) Magnetic Intensity ($H$) and (ii) Intensity of Magnetization ($M_m$). State their SI units.
Answer:
(i) Magnetic Intensity ($H$): It is the capability of an external magnetic field to magnetize a magnetic material.
(ii) Intensity of Magnetization ($M_m$): It is defined as the magnetic dipole moment developed per unit volume of a material when placed in an external magnetizing field.
Both share the same SI unit: $\text{A/m}$ (Ampere per meter).
(i) Magnetic Intensity ($H$): It is the capability of an external magnetic field to magnetize a magnetic material.
(ii) Intensity of Magnetization ($M_m$): It is defined as the magnetic dipole moment developed per unit volume of a material when placed in an external magnetizing field.
Both share the same SI unit: $\text{A/m}$ (Ampere per meter).
Long Answer Questions (5 Marks)
9. Derive expressions for the torque and potential energy of a magnetic dipole placed in a uniform magnetic field. Mention the conditions for stable and unstable equilibrium.
Answer Hint: Consider a bar magnet of length $2l$ and pole strength $m$ at an angle $\theta$ in field $B$.
Force on N-pole = $+mB$, Force on S-pole = $-mB$.
Torque $\tau = \text{Force} \times \text{perpendicular distance} = mB \times 2l\sin\theta = MB\sin\theta$.
Work done $dW = \tau d\theta = MB\sin\theta d\theta$. Integrate from $\theta_1 = 90^\circ$ to $\theta_2 = \theta$ to obtain $U = -MB\cos\theta$.
Detail stable ($\theta=0^\circ$) and unstable ($\theta=180^\circ$) states as shown in the concepts section.
Force on N-pole = $+mB$, Force on S-pole = $-mB$.
Torque $\tau = \text{Force} \times \text{perpendicular distance} = mB \times 2l\sin\theta = MB\sin\theta$.
Work done $dW = \tau d\theta = MB\sin\theta d\theta$. Integrate from $\theta_1 = 90^\circ$ to $\theta_2 = \theta$ to obtain $U = -MB\cos\theta$.
Detail stable ($\theta=0^\circ$) and unstable ($\theta=180^\circ$) states as shown in the concepts section.
10. Describe the classification of magnetic materials into diamagnetic, paramagnetic, and ferromagnetic materials on the basis of behavior in non-uniform fields, relative permeability, and magnetic susceptibility.
Answer Hint: Explain that:
1. Diamagnetic substances move from stronger to weaker parts of a non-uniform field, have negative $\chi_m$, and $\mu_r < 1$.
2. Paramagnetic substances move feebly from weaker to stronger parts, have positive small $\chi_m$, and $\mu_r > 1$.
3. Ferromagnetic substances move strongly from weaker to stronger parts, have exceptionally high positive $\chi_m$, and $\mu_r \gg 1$.
1. Diamagnetic substances move from stronger to weaker parts of a non-uniform field, have negative $\chi_m$, and $\mu_r < 1$.
2. Paramagnetic substances move feebly from weaker to stronger parts, have positive small $\chi_m$, and $\mu_r > 1$.
3. Ferromagnetic substances move strongly from weaker to stronger parts, have exceptionally high positive $\chi_m$, and $\mu_r \gg 1$.
Assertion-Reason Questions
(Options: A = Both True & R is correct explanation, B = Both True & R is not correct, C = A is True & R is False, D = A is False & R is True)
11. Assertion: The net magnetic flux through any closed surface is always zero.
Reason: Isolated magnetic poles (monopoles) do not exist in nature.
Answer: (A) Both are true, and the reason is the correct explanation of the assertion.
12. Assertion: When a ferromagnetic material is heated beyond its Curie temperature, it transitions into a paramagnetic material.
Reason: Thermal agitation disrupts the alignment of magnetic domains within the material.
Answer: (A) Both are true, and the reason perfectly explains why domains break up at high temperatures.
Case-Based Questions
13. Magnetic Field Lines Analysis:
Magnetic field lines are visual tools used to represent the distribution of a magnetic field. They follow specific rules: they never intersect, their density indicates field strength, and they form closed paths.
Q1. Why can two magnetic field lines never cross each other?
Answer: If they intersect, there would be two different directions of the magnetic field at the point of intersection, which is physically impossible.
Q2. What is the direction of magnetic field lines inside a bar magnet?
Answer: Inside the bar magnet, the field lines travel from the South Pole to the North Pole.
Answer: If they intersect, there would be two different directions of the magnetic field at the point of intersection, which is physically impossible.
Q2. What is the direction of magnetic field lines inside a bar magnet?
Answer: Inside the bar magnet, the field lines travel from the South Pole to the North Pole.
14. Materials under External Magnetic Fields:
When materials are placed in an external magnetic field, their atomic dipoles respond differently depending on their atomic structure.
Q1. Which material expels magnetic lines of force?
Answer: Diamagnetic materials expel lines of force because they develop internal magnetization opposite to the external field.
Q2. Name one example each of a diamagnetic and a ferromagnetic material.
Answer: Diamagnetic: Copper (or Water/Bismuth). Ferromagnetic: Iron (or Cobalt/Nickel).
Answer: Diamagnetic materials expel lines of force because they develop internal magnetization opposite to the external field.
Q2. Name one example each of a diamagnetic and a ferromagnetic material.
Answer: Diamagnetic: Copper (or Water/Bismuth). Ferromagnetic: Iron (or Cobalt/Nickel).
Common Mistakes Students Make
- Direction of Dipole Moment vs. Field Lines: Remember that the magnetic dipole moment vector ($\mathbf{M}$) points from South to North, whereas external magnetic field lines travel from North to South. Don't swap them!
- Sign of Work Done: Pay close attention to the formula $W = MB(\cos\theta_1 - \cos\theta_2)$. If you turn a magnet from $0^\circ$ to $180^\circ$, it becomes $\cos(0^\circ) - \cos(180^\circ) = 1 - (-1) = 2$. Students often write $1-1=0$ by mistake.
- Susceptibility Unit Confusion: Magnetic susceptibility ($\chi_m$) is a pure ratio of identical quantities ($M_m/H$). It is completely dimensionless and has no units! Do not write $\text{A/m}$ for it.
Exam Preparation Tips
- Focus on the Tables: Learn the comparison table between Dia, Para, and Ferro materials. One question from this table is practically guaranteed every year.
- Practice Core Derivations: Make sure you can comfortably derive the expression for the torque and potential energy of a dipole.
- Keep Core Definitions Clear: Write precise definitions for magnetic intensity, relative permeability, and susceptibility to score full marks in short-answer questions.
- Solve Numerical Step-by-Step: Write down the given variables first, state the formula explicitly, and carry out the calculations with correct SI units. This ensures step-marking rewards even if your final arithmetic goes wrong.
FAQ Section
Is Earth's Magnetism included in the CBSE Class 12 Physics syllabus for 2026?
No, topics like Earth's magnetic elements, magnetic declination, angle of dip, and horizontal component have been removed from the current rationalized NCERT syllabus. Focus on bar magnets and magnetic properties instead.
Where can I download the updated NCERT solutions PDF?
You can access all the latest updated chapter-wise solutions, worksheets, and study notes directly on examspark.in.
What is the unit of magnetic dipole moment?
The SI unit of magnetic dipole moment is $\text{A m}^2$ (Ampere meter-squared) or $\text{J/T}$ (Joules per Tesla).
What happens to a ferromagnetic material at Curie Temperature?
At the Curie temperature, a ferromagnetic material loses its strong domain-based magnetic properties and transitions completely into a paramagnetic material due to thermal disruption.
Why is Gauss's Law in magnetism different from electrostatics?
In electrostatics, net flux can be non-zero because isolated positive or negative charges can exist. In magnetism, poles always exist in pairs, so any line leaving a surface must eventually re-enter it, making the net flux zero.
Chapter 5, Magnetism and Matter, is highly conceptual and completely straightforward to master if you keep your fundamentals clear. Ratta mat maro (avoid blind memorization)-focus on the core distinctions between the types of matter and how a bar magnet interacts with external fields. Prepare for your 2026 board exams with zero stress! Keep practicing your past year questions (PYQs), cross-verify your answers with these solutions, and download your quick reference study sheets on examspark.in. Stay consistent and study hard!