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NCERT Exercises and Solutions: Magnetism and Matter

🎓 Class 12 Physics CBSE Theory Ch 5 – Magnetism and Matter ⏱ ~8 min
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NCERT Exercises and Solutions: Magnetism and Matter

Chapter 5 - Summary at a Glance

This chapter introduced the magnetism of matter and the magnetism of the Earth itself. The key results to remember:

Bar magnet as a dipole. Pole strength qm, magnetic length 2l, dipole moment m = qm(2l). On the axis at distance r:
\(B_{axial} = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}\)
On the equator: \(B_{eq} = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\) ⇒ axial = 2 × equatorial.
Torque and energy. τ = m × B; U = −m·B. Stable at θ = 0; unstable at θ = 180°.
Gauss's law for magnetism. \(\oint \vec B\cdot d\vec A = 0\) - magnetic monopoles do not exist.
Geomagnetic elements. Three numbers fully specify Earth's field at any place: declination D, dip I, horizontal component H. BE = H/cos I, Z = BE sin I.
Magnetisation and intensity. M = mnet/V; H = B/μ0 − M ⇒ B = μ0(H + M). For linear media, M = χH, B = μ0(1 + χ)H = μH; μr = 1 + χ.
Diamagnetism. χ small and negative; M opposes H. Examples: Bi, Cu, H₂O. χ is independent of T. Superconductors: perfect diamagnets, χ = −1.
Paramagnetism. χ small and positive; M aligns weakly with H. χ ∝ 1/T (Curie's law). Examples: Al, Pt, O₂.
Ferromagnetism. χ very large; spontaneous domain alignment below Curie temperature TC. Hysteresis loop. Soft ferromagnets (low coercivity, narrow loop): transformer cores. Hard ferromagnets (high retentivity & coercivity): permanent magnets.

Key Terms

TermDefinition
Magnetic dipole moment mqm(2l) for bar magnet; NIA for current loop. Unit: A m².
Magnetisation MNet dipole moment per unit volume (A/m).
Magnetic intensity HB/μ₀ − M (A/m); set by free currents.
Susceptibility χM/H (dimensionless); classifies materials.
Permeability μμ₀(1 + χ); μr = μ/μ₀.
Declination DAngle between magnetic and geographic north.
Dip / Inclination IAngle B makes with horizontal.
Curie temperature TCAbove this T, ferromagnet → paramagnet.
Retentivity BrB remaining when H is reduced to zero.
Coercivity HcReverse H needed to bring B to zero.

NCERT Exercises - Worked Solutions

Exercise 5.1 L2 Understand

Answer the following:

(a) A vector needs three quantities for its specification. Name the three independent conventional quantities used for specifying the Earth's magnetic field.

(b) The angle of dip at a place is greater in Britain (50° N latitude) than in southern India. Why?

(c) If you made a map of magnetic field lines of Mumbai's bar magnet, would the lines be open or closed?

(d) If a bar magnet is cut into two equal halves perpendicular to its axis, what happens to its dipole moment?

(a) Magnetic declination D, magnetic dip (inclination) I, horizontal component H.

(b) Higher (magnetic) latitude ⇒ closer to magnetic pole ⇒ field is more vertical ⇒ greater dip. Britain at ~70° dip; south India ~10°.

(c) Closed - magnetic field lines always form closed loops because monopoles don't exist (Gauss's law for magnetism).

(d) Each half is a complete magnet but with the same length 2l/2 = l. New pole strength = qm; new moment m' = qm × (l) = m/2. So dipole moment halves.

Exercise 5.2 L3 Apply

A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm. (a) What is the magnetic moment of the magnet? (b) What is the work done in moving it from its most stable to its most unstable position? (c) The bar magnet is replaced by a solenoid of cross-section 2 × 10⁻⁴ m² and 1000 turns; what is the current required to give the same magnetic moment?

τ = mB sin θ ⇒ m = τ/(B sin θ) = 0.016/(800 × 10⁻⁴ × sin 30°) = 0.016/(0.08 × 0.5) = 0.40 A m².

(b) Work done = U(180°) − U(0°) = mB − (−mB) = 2mB = 2 × 0.40 × 0.08 = 0.064 J.

(c) m = NIA ⇒ I = m/(NA) = 0.40/(1000 × 2 × 10⁻⁴) = 2.0 A.

Exercise 5.3 L3 Apply

A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

The field outside a current-carrying solenoid is essentially identical to that of a bar magnet; one face acts as N, the other as S, by the right-hand rule.

m = NIA = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.60 A m².

Exercise 5.4 L3 Apply

If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

τ = mB sin θ = 0.60 × 0.25 × sin 30° = 0.60 × 0.25 × 0.5 = 0.075 N m.

Exercise 5.5 L3 Apply

A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its moment (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?

(a) U(θ) = −mB cos θ. (i) W = U(90°) − U(0°) = 0 − (−mB) = mB = 1.5 × 0.22 = 0.33 J. (ii) W = U(180°) − U(0°) = mB − (−mB) = 2mB = 0.66 J.

(b) τ = mB sin θ. (i) θ = 90°: τ = 1.5 × 0.22 × 1 = 0.33 N m. (ii) θ = 180°: τ = 0.

Exercise 5.6 L3 Apply

The horizontal component of the Earth's magnetic field at a certain place is 3.0 × 10⁻⁵ T and the angle of dip is 30°. Find the strength of the Earth's magnetic field at the place.

H = BE cos I ⇒ BE = H/cos I = 3.0 × 10⁻⁵/cos 30° = 3.0 × 10⁻⁵/(0.866) ≈ 3.46 × 10⁻⁵ T.

Exercise 5.7 L4 Analyse

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 centre of the magnet on (a) the axis, (b) the equatorial line of the magnet.

μ₀/4π = 10⁻⁷ T m/A; r = 0.10 m; m = 0.48 A m².

(a) Baxial = 10⁻⁷ × 2 × 0.48/(0.1)³ = 9.6 × 10⁻⁵ T, along m (S → N).

(b) Beq = 10⁻⁷ × 0.48/(0.1)³ = 4.8 × 10⁻⁵ T, opposite to m.

Exercise 5.8 L4 Analyse

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre. The Earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector at the same distance? (At null point, the field due to the magnet is equal and opposite to HE.)

At null point on axis: Baxial = HE = 0.36 G (since dip = 0, H = BE).

On equator at the same r: Beq = ½ Baxial = 0.18 G, but pointing opposite to m.

Earth's field at that point still = 0.36 G along magnet axis (geographic N). Net field = HE − Beq in same line ⇒ |Btotal| = 0.36 + 0.18 = 0.54 G (since both add along the equatorial perpendicular bisector convention).

Answer: 0.54 G = 5.4 × 10⁻⁵ T.

Exercise 5.9 L4 Analyse

If the bar magnet in exercise 5.8 is turned around by 180°, where will the new null points be located?

After rotating, the axial field of the magnet now points opposite to HE. Null point now occurs on the equator (which used to be the axis direction), where Beq = HE ⇒ Beq = (μ₀/4π)(m/r'³) = HE.

Compare with original: HE = (μ₀/4π)(2m/r³). Hence r'³ = r³/2 ⇒ r' = r × (1/2)^(1/3) = 14 × 0.794 ≈ 11.1 cm.

Null points are now ~ 11.1 cm from the centre on the equatorial line (i.e. east-west direction).

Exercise 5.10 L3 Apply

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the Earth's magnetic field is 0.35 G. Determine the magnitude of the Earth's magnetic field at the place.

The 22° below horizontal is the angle of dip I. BE = H/cos I = 0.35/cos 22° = 0.35/0.927 ≈ 0.38 G.

Exercise 5.11 L4 Analyse

At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the Earth's field is measured to be 0.16 G. Specify the direction and magnitude of the Earth's field at the location.

D = 12° west, I = 60°, H = 0.16 G. BE = H/cos I = 0.16/cos 60° = 0.16/0.5 = 0.32 G.

The field points 12° west of geographic north in horizontal projection and dips at 60° below horizontal toward the magnetic-north direction.

Exercise 5.12 L4 Analyse

A short bar magnet has magnetic moment 0.36 J/T and is placed in vacuum. (a) Find the magnetic field on the axis at 20 cm from the centre. (b) Find the field on the equator at the same distance.

r = 0.20 m. (a) Baxial = 10⁻⁷ × 2 × 0.36/(0.20)³ = 9.0 × 10⁻⁶ T.

(b) Beq = 10⁻⁷ × 0.36/(0.20)³ = 4.5 × 10⁻⁶ T.

Interactive: Mixed Numerical Practice L3 Apply

Slide to set m and r and read off both fields, exactly as in exercises 5.7 and 5.12.

Baxial = 9.0 × 10⁻⁶ T  |  Beq = 4.5 × 10⁻⁶ T

Competency-Based Questions L1-L6

A physics teacher demonstrates: a soft-iron rod placed inside a solenoid increases B 2000-fold; a copper rod barely changes it.
1. The soft-iron rod is __ ; the copper rod is __ . L1 Remember
  • (a) ferromagnetic; diamagnetic
  • (b) paramagnetic; ferromagnetic
  • (c) diamagnetic; ferromagnetic
  • (d) ferromagnetic; paramagnetic
(a).
2. Why is the soft-iron core inside an electromagnet able to switch on and off cleanly with the current? L2 Understand
Soft iron has very low retentivity; once H drops to zero, B nearly drops to zero too. Domains relax to a random configuration almost immediately.
3. State two factors on which the magnetic moment of a current loop depends. L1 Remember
m = NIA depends on the number of turns N (and the loop area A) and the current I.
4. Compare the temperature dependence of χ for a paramagnetic and a diamagnetic material. L4 Analyse
Paramagnet: χ ∝ 1/T (Curie's law) - falls as T rises. Diamagnet: χ is essentially independent of T because it arises from induced orbital perturbation, not thermal alignment.
5. Propose a project to magnetise a steel knitting needle and use it as a compass needle. List materials, steps and a verification test. L6 Create
Materials: steel needle, strong bar magnet, cork, dish of water, compass for verification. Steps: stroke the needle 30 times in one direction with one pole of the magnet (single-stroke method) - this creates a residual M. Float the needle on a piece of cork in the water dish; it will line up north-south. Verification: compare with the direction shown by the lab compass; or push a paper clip near each end - the freshly magnetised tip will attract the clip.

Assertion-Reason Pairs L4 Analyse

Options: (A) Both true, R correct explanation. (B) Both true, R not the explanation. (C) A true, R false. (D) A false, R true.

Assertion: When a bar magnet is cut along its length, each piece has weaker pole strength.
Reason: Cutting along the length halves the cross-sectional area, so each half encloses fewer aligned dipoles.
(A). True; halving the area halves the number of aligned domains crossing the polar face, hence pole strength.
Assertion: The dip needle is horizontal at the magnetic equator.
Reason: At the magnetic equator, the Earth's field is horizontal (I = 0).
(A). Reason directly explains the assertion.
Assertion: The relative permeability of a ferromagnet is constant.
Reason: M = χH is a linear relation for ferromagnets.
(D). Both statements are false: ferromagnets are nonlinear and μr varies with H (and history).

Frequently Asked Questions - NCERT Exercises and Solutions: Magnetism and Matter

What are the key NCERT exercise types in Chapter 5 Magnetism and Matter?
NCERT Class 12 Physics Chapter 5 Magnetism and Matter exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Magnetism and Matter?
For numerical problems in NCERT Class 12 Physics Chapter 5 Magnetism and Matter: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 5?
From NCERT Class 12 Physics Chapter 5 (Magnetism and Matter), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 5 Magnetism and Matter problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 5 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 5 Magnetism and Matter exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 5 Magnetism and Matter solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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