Chapter 5 Magnetism and Matter

NCERT Solutions Class 12 Physics Chapter 5 - Magnetism and Matter

In 12th standard Physics, Chapter 5 Magnetism and Matter builds upon the topic of magnetism and how materials respond to the earth's magnetic field, which was introduced in the previous chapter on the phenomenon of current producing magnetism. This chapter also addresses the concept of the earth as a large magnet, categorising materials into paramagnetic, diamagnetic and ferromagnetic based on the degree of their response to an external magnetic field. This chapter is one of the higher marks achieved in the CBSE Board and the JEE and NEET exams, as it relates to a deeper understanding of the microscopic basis of magnetism.

NCERT Solutions for Class 12 Physics Chapter 5 (Magnetism and Matter) have been developed to assist students with a better understanding of the earth's magnetic field (and the vector components of the magnetic field) and the qualitative characteristics of different types of magnetic materials. NCERT Class 12 Physics Chapter 5 Solutions also aid students in understanding the differences between the electric dipole and magnetic dipole and allow students to transition from studying electric systems in the prior chapters of physics to studying magnetic systems.

1.0Class 12 Physics Chapter 5 Magnetism and Matter: Key Concepts

Class 12 Physics Chapter 5, Magnetism and Matter, explains the properties of magnets, the Earth’s magnetic field, and how different materials behave in a magnetic field. The main concepts covered in this chapter include:

Bar Magnet: Understand the magnetic field produced by a bar magnet and its similarity with a current-carrying solenoid. The chapter also explains how a bar magnet behaves when placed in an external magnetic field.
Gauss’s Law of Magnetism: Learn that magnetic monopoles do not exist and that the net magnetic flux through any closed surface is zero.
Earth’s Magnetic Field: Study the magnetic field of the Earth, including the magnetic poles and magnetic equator. The chapter also covers magnetic declination, angle of dip, and the horizontal component of the Earth’s magnetic field.
Magnetisation and Magnetic Intensity: Understand how materials respond to an external magnetic field. The chapter introduces magnetic susceptibility and permeability and explains their relation to the magnetic properties of materials.
Types of Magnetic Materials: Learn about diamagnetic, paramagnetic, and ferromagnetic materials based on their behaviour in an external magnetic field. The chapter also explains the arrangement of magnetic dipoles in these materials.
Hysteresis: Understand the lag of magnetisation behind the magnetising field in ferromagnetic materials. The hysteresis curve also helps explain the choice of materials for making permanent magnets and electromagnets.

2.0NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter : Detailed Solutions

SOLVED EXAMPLES

(a) What happens if a bar magnet is cut into two pieces : (i) transverse to its length, (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B]

Sol.

(a) In either case, one gets two magnets, each with a north and south-pole. (i) Pole strength same (ii) Pole strength becomes half. (b) No force if the field is uniform. The iron nail experiences a non-uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south-pole (say) in the nail is closer to the north pole of magnet than induced north pole. (c) Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor. (d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is a magnet, pick up one, (say, A) and lower one of its ends; first on one of the ends of the other (say, B), and then on the middle of B. If you notice that in the middle of B, A experiences less force, then B is magnetised. If you do not notice any change from the end to the middle of B, then A is magnetised.

Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration the system is not in equilibrium? (b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown?

Circular Gaussian path enclosing charges Q1 and Q2, with external charges Q3 to Q6 and point P shown inside

Sol. Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression (Eq.)

$B_{P} = - \frac{μ _{0}}{4 π} \frac{M _{P}}{r ^{3}} B_{P} = \frac{μ _{0} 2}{4 π} \frac{M _{P}}{r ^{3}} (on the normal bisector) (on the axis)$

where $M_{p}$ is the magnetic moment of the dipole P. Equilibrium is stable when $M_{Q}$ is parallel to $B_{P}$ , and unstable when it is anti-parallel to $B_{P}$ . For instance for the configuration $Q_{3}$ for which Q is along the perpendicular bisector of the dipole, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence $Q_{3}$ is stable. Thus, (a) $PQ_{1}$ and $PQ_{2}$ (b) (i) $PQ_{3}, PQ_{6}$ (stable) (ii) $PQ_{5}, PQ_{4}$ (unstable) (c) $PQ_{6}$

Many of the diagrams given in figure show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.

Diagrams illustrating different magnetic field patterns, including a magnetic pole, uniform field, circular fields, solenoid, bar magnets, and toroidal magnetic field.

Sol.

(a) Wrong : Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor.

(b) Wrong : Magnetic field lines (like electric field lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges. (c) Right : Magnetic lines are completely confined within a toroid. Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field. (d) Wrong : Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere's law. The lines should curve out at both ends, and meet eventually to form closed loops. (e) Right : These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the net flux of the field is zero. (f) Wrong : These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. The difference between Fig. [(e) and (f)] should be carefully grasped. (g) Wrong : Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere's law is violated. This is also true for electric field lines.

(a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point? (b) If magnetic monopoles existed, how would the Gauss's law of magnetism be modified? (c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire? (d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero?

Sol.

(a) No. The magnetic force is always normal to B [remember magnetic force $= q (v \times B)$ ]. It is misleading to call magnetic field lines as lines of force. (b) Gauss's law of magnetism states that the flux of $B$ -through any closed surface is always zero.

$\int_{s} B \cdot Δ s = 0$

If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) $q_{m}$ enclosed by S. [Analogous to Gauss's law of electrostatics, $\int_{s} B .Δ s = μ_{0} q_{m}$ where $q_{m}$ is the (monopole) magnetic charge enclosed by S.] (c) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.) (d) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.

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 2 A . If the number of turns is 1000 per metre. Calculate (a) H. (b) Magnetisation (I) (c) B and (d) the magnetising current $i_{m}$ . Sol. (a) The field H is dependent of the material of the core and is,

$H = ni = 1000 \times 2.0 = 2 \times 1 0^{3} A / m .$

(b) The magnetic field B is given by

$B = μ_{r} μ_{0} H = 400 \times 4 π \times 1 0^{- 7} (N / A^{2}) \times 2 \times 1 0^{3} (A / m) = 1.0 T$

$I = (B - μ_{0} H) / μ_{0} = (μ_{r} μ_{0} H - μ_{0} H) / μ_{0} = (μ_{r} - 1) H = 399 \times H ≅ 8 \times 1 0^{5} A / m$

(d) The magnetising current $i_{m}$ is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus,

$B = μ_{r} μ_{0} n (i + i_{m})$

Using $i = 2 A, B = 1 T$ , We get $i_{m} = 794 A$ .

EXERCISE QUESTION WITH SOLUTIONS

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5 \times 1 0^{- 2} J$ . What is the magnitude of magnetic moment of the magnet?

Sol. $B = 0.25 T, τ = 4.5 \times 1 0^{- 2} J, θ = 3 0^{\circ}$ Torque on magnet,

$∴ \Rightarrow τ = MB sin θ M = \frac{τ}{B sin θ} M = \frac{4.5 \times 1 0 ^{- 2}}{0.25 \times sin 3 0 ^{\circ}} = 0.36 J / T$

A short bar magnet of magnetic moment $M = 0.32 J / T$ is placed in a uniform magnetic field of 0.15 T. If the bar 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?

Sol. Here; $M = 0.32 J / T, B = 0.15 T$ We know that,

$U = - M \cdot B = - MB cos θ$

(a) For stable equilibrium, $M$ is to be parallel to $B$ . In this case, $θ = 0^{\circ}$ . Thus,

$U = - MB cos 0^{\circ} = - MB = - 0.32 \times 0.15 = - 4.8 \times 1 0^{- 2} J$

(b) For unstable equilibrium, $M$ is to be antiparallel to $B$ . in this case, $θ = 18 0^{\circ}$ . Thus,

$U = - MB cos 18 0^{\circ} = - MB (- 1) = MB = 0.32 \times 0.15 = 4.8 \times 1 0^{- 2} J$

A closely wound solenoid of 800 turns and area of cross section $2.5 \times 1 0^{- 4} m^{2}$ 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 ? Sol. Here; $N = 800, i = 3.0 A, A = 2.5 \times 1 0^{- 4} m^{2}$ Magnetic moment of solenoid,

$M = NiA = 800 \times 3 \times 2.5 \times 1 0^{- 4} = 0.60 A - m^{2}, along axis of solenoid from$

south to north pole. Field lines pattern of solenoid is equivalent to field lines pattern of bar magnet.

If the solenoid in question 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? Sol. $B = 0.25 T, M = 0.6 A - m^{2}$ The angle $θ$ , between the axis of the solenoid and the direction of the applied field is 30°. Therefore, the torque acting on the solenoid,

$τ = MB sin θ = 0.6 \times 0.25 sin 3 0^{\circ} = 7.5 \times 1 0^{- 2} N - m$

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 magnetic 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)? Sol. (a) Magnetic moment, $M = 1.5 J / T$ , $B = 0.22 T$ , (i) The work required to make the magnetic moment normal to the direction of magnetic field,

$W = MB (cos θ_{1} - cos θ_{2}) θ_{1} = 0^{\circ}, θ_{2} = 9 0^{\circ}] = 1.5 \times 0.22 (cos 0^{\circ} - cos 9 0^{\circ}) = 0.33 (1 - 0) = 0.33 J$

(ii) $θ_{1} = 0^{\circ}, θ_{2} = 18 0^{\circ}$ The work required to make the magnetic moment opposite to the direction of magnetic field,

$W = MB (cos θ_{1} - cos θ_{2}) = 1.5 \times 0.22 (cos 0^{\circ} - cos 18 0^{\circ}) = 0.33 [1 - (- 1)] = 0.66 J$

(b) For case (i) : $θ = θ_{2} = 9 0^{\circ}$

$∴ Torque, τ = MB sin θ = 1.5 \times 0.22 sin 9 0^{\circ} = 0.33 N - m$

For case (ii) : $θ = θ_{2} = 18 0^{\circ}$

$∴ Torque, τ = MB sin θ = MB sin 18 0^{\circ} = 0 N - m$

A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 1 0^{- 4} m^{2}$ , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 1 0^{- 2} T$ is set up at an angle of 30° with the axis of the solenoid?

Sol. $N = 2000, A = 1.6 \times 1 0^{- 4} m^{2}, I = 4 A$

(a) The magnetic moment along the axis of the solenoid,

$M = NIA = 2000 \times 4 \times 1.6 \times 1 0^{- 4} = 1.28 Am^{2}$

(b) The net force experienced by a magnetic dipole in a uniform magnetic field is zero. The torque $(τ)$ on the solenoid is given by,

$τ = MB sin θ = 1.28 \times 7.5 \times 1 0^{- 2} sin 3 0^{\circ} = 4.8 \times 1 0^{- 2} Nm$

The torque tends to align the axis of the solenoid (i.e., its magnetic moment vector $M$ ) parallel to the magnetic field $B$ .

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 lines (normal bisector) of the magnet. Sol. $M = 0.48 J / T, r = 10 cm = 0.1 m$ (a) The magnetic field at distance $r$ , from the centre of the magnet on the axis,

$∴ = B_{axis} = \frac{μ _{0}}{4 π} \frac{2 M}{r ^{3}} B_{axis} = \frac{4 π \times 1 0 ^{- 7} \times 2 \times 0.48}{4 π \times ( 0.1 ) ^{3}} 0.96 \times 1 0^{- 4} T = 0.96 G$

[The magnetic field is along the S-N direction] (b) The magnetic field at a distance of 10 cm on the equatorial line of the magnet

$B_{equatorial} = \frac{μ _{0} \times M}{4 π \times r ^{3}} = \frac{4 π \times 1 0 ^{- 7} \times 0.48}{4 π ( 0.1 ) ^{3}} = 0.48 G$

[The magnetic field is along the N-S direction]

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 of the magnet. 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 of the magnet at the same distance as the null-point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.) Sol. As the null point $N_{1}$ and $N_{2}$ lie on the axis (S-N) of the bar magnet, its south pole points towards the magnetic north pole of the earth as dip is zero. Clearly, for points $N_{1}$ and $N_{2}, B_{H}$ and $B_{axial}$ are in opposite direction and

$or B_{axial} = B_{H} \frac{μ _{o}}{4 π} \frac{2 M}{r ^{3}} = B_{H} \frac{μ _{o}}{4 π} \frac{2 M}{r ^{3}} = 0.36 G$

Bar magnet showing axial and equatorial magnetic field components relative to Earth’s horizontal magnetic field.

Magnetic field at a point P on the normal bisector (i.e., equatorial line) of the magnet at the same distance (r) from its centre O, i.e.,

$B_{equatorial} = \frac{μ _{0}}{4 π} \frac{M}{r ^{3}} = \frac{B _{H}}{2} B_{equatorial} = \frac{0.36 G}{2} = 0.18 G$

Since $B_{H}$ and $B_{equatorial}$ are in the same direction at point P, so total magnetic field at

$P, B = B_{H} + B_{equatorial} = 0.36 G + 0.18 G = 0.54 G .$

If the bar magnet in question 8 is turned round by $18 0^{\circ}$ , where will the new null points be located? Sol. When the magnet is turned around by 180°, its south pole will point toward magnetic south of the Earth [fig.] clearly, there are two null points ( $N_{1}$ and $N_{2}$ ) on normal bisector. The null point lie on the equatorial line of the magnet at a distance $r^{'}$ , from the centre O of the magnet, thus for points $N_{1}$ and $N_{2} -$ There are two null points on normal bisector $B_{H} = B_{equatorial}$ (in the plane of paper)

Bar magnet with axial and equatorial magnetic field components shown at symmetric points on either side.

$B_{H} = \frac{μ _{o}}{4 π} \frac{M}{( r ^{'} ) ^{3}} \Rightarrow (r^{'})^{3} = \frac{μ _{o}}{4 π} \frac{M}{B _{H}} r^{'} = (\frac{μ _{o}}{4 π} \frac{M}{B _{H}})^{1/3},$

where $\frac{μ _{o}}{4 π} \frac{M}{B _{H}} = \frac{r ^{3}}{2}, r = 14 cm$ (from question 8)

$r^{'} = (\frac{r ^{3}}{2})^{1/3} = 14 \times 2^{- 1/3} = 11.1 cm$

from the centre of bar magnet.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for Class 12 Physics for all chapters, with exercise answers, important formulas, and clear solutions to help students understand concepts and practise questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 5

Comparison of Magnetic Materials: The solutions clearly compare diamagnetic, paramagnetic, and ferromagnetic materials, including their important properties and behaviour in a magnetic field.
Clear Explanation of Earth’s Magnetic Field: The solutions explain the magnetic meridian, geographic meridian, angle of dip, and horizontal component of the Earth’s magnetic field to help students understand related questions.
Bar Magnet and Solenoid: The solutions explain the similarity between a bar magnet and a current-carrying solenoid and help students understand the magnetic field produced by both.
Understanding Hysteresis: Retentivity and coercivity are explained step by step. The solutions also explain why soft iron is used for electromagnets and transformer cores, while materials with high retentivity are suitable for permanent magnets.
Clear Definitions and Units: Important quantities such as magnetic moment, magnetisation, magnetic intensity, susceptibility, and permeability are explained with their definitions and units to help students solve numerical questions correctly.

Frequently Asked Questions

NCERT Solutions Class 12 Physics Chapter 5 - Magnetism and Matter

1.0Class 12 Physics Chapter 5 Magnetism and Matter: Key Concepts

Bar Magnet: Understand the magnetic field produced by a bar magnet and its similarity with a current-carrying solenoid. The chapter also explains how a bar magnet behaves when placed in an external magnetic field.
Gauss’s Law of Magnetism: Learn that magnetic monopoles do not exist and that the net magnetic flux through any closed surface is zero.
Earth’s Magnetic Field: Study the magnetic field of the Earth, including the magnetic poles and magnetic equator. The chapter also covers magnetic declination, angle of dip, and the horizontal component of the Earth’s magnetic field.
Magnetisation and Magnetic Intensity: Understand how materials respond to an external magnetic field. The chapter introduces magnetic susceptibility and permeability and explains their relation to the magnetic properties of materials.
Types of Magnetic Materials: Learn about diamagnetic, paramagnetic, and ferromagnetic materials based on their behaviour in an external magnetic field. The chapter also explains the arrangement of magnetic dipoles in these materials.
Hysteresis: Understand the lag of magnetisation behind the magnetising field in ferromagnetic materials. The hysteresis curve also helps explain the choice of materials for making permanent magnets and electromagnets.

2.0NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter : Detailed Solutions

SOLVED EXAMPLES

(a) What happens if a bar magnet is cut into two pieces : (i) transverse to its length, (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B]

Sol.

Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration the system is not in equilibrium? (b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown?

$B_{P} = - \frac{μ _{0}}{4 π} \frac{M _{P}}{r ^{3}} B_{P} = \frac{μ _{0} 2}{4 π} \frac{M _{P}}{r ^{3}} (on the normal bisector) (on the axis)$

Many of the diagrams given in figure show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.

Sol.

(a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point? (b) If magnetic monopoles existed, how would the Gauss's law of magnetism be modified? (c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire? (d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero?

Sol.

$\int_{s} B \cdot Δ s = 0$

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 2 A . If the number of turns is 1000 per metre. Calculate (a) H. (b) Magnetisation (I) (c) B and (d) the magnetising current $i_{m}$ . Sol. (a) The field H is dependent of the material of the core and is,

$H = ni = 1000 \times 2.0 = 2 \times 1 0^{3} A / m .$

(b) The magnetic field B is given by

$B = μ_{r} μ_{0} H = 400 \times 4 π \times 1 0^{- 7} (N / A^{2}) \times 2 \times 1 0^{3} (A / m) = 1.0 T$

$I = (B - μ_{0} H) / μ_{0} = (μ_{r} μ_{0} H - μ_{0} H) / μ_{0} = (μ_{r} - 1) H = 399 \times H ≅ 8 \times 1 0^{5} A / m$

$B = μ_{r} μ_{0} n (i + i_{m})$

Using $i = 2 A, B = 1 T$ , We get $i_{m} = 794 A$ .

EXERCISE QUESTION WITH SOLUTIONS

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5 \times 1 0^{- 2} J$ . What is the magnitude of magnetic moment of the magnet?

Sol. $B = 0.25 T, τ = 4.5 \times 1 0^{- 2} J, θ = 3 0^{\circ}$ Torque on magnet,

$∴ \Rightarrow τ = MB sin θ M = \frac{τ}{B sin θ} M = \frac{4.5 \times 1 0 ^{- 2}}{0.25 \times sin 3 0 ^{\circ}} = 0.36 J / T$

A short bar magnet of magnetic moment $M = 0.32 J / T$ is placed in a uniform magnetic field of 0.15 T. If the bar 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?

Sol. Here; $M = 0.32 J / T, B = 0.15 T$ We know that,

$U = - M \cdot B = - MB cos θ$

(a) For stable equilibrium, $M$ is to be parallel to $B$ . In this case, $θ = 0^{\circ}$ . Thus,

$U = - MB cos 0^{\circ} = - MB = - 0.32 \times 0.15 = - 4.8 \times 1 0^{- 2} J$

(b) For unstable equilibrium, $M$ is to be antiparallel to $B$ . in this case, $θ = 18 0^{\circ}$ . Thus,

$U = - MB cos 18 0^{\circ} = - MB (- 1) = MB = 0.32 \times 0.15 = 4.8 \times 1 0^{- 2} J$

A closely wound solenoid of 800 turns and area of cross section $2.5 \times 1 0^{- 4} m^{2}$ 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 ? Sol. Here; $N = 800, i = 3.0 A, A = 2.5 \times 1 0^{- 4} m^{2}$ Magnetic moment of solenoid,

$M = NiA = 800 \times 3 \times 2.5 \times 1 0^{- 4} = 0.60 A - m^{2}, along axis of solenoid from$

south to north pole. Field lines pattern of solenoid is equivalent to field lines pattern of bar magnet.

If the solenoid in question 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? Sol. $B = 0.25 T, M = 0.6 A - m^{2}$ The angle $θ$ , between the axis of the solenoid and the direction of the applied field is 30°. Therefore, the torque acting on the solenoid,

$τ = MB sin θ = 0.6 \times 0.25 sin 3 0^{\circ} = 7.5 \times 1 0^{- 2} N - m$

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 magnetic 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)? Sol. (a) Magnetic moment, $M = 1.5 J / T$ , $B = 0.22 T$ , (i) The work required to make the magnetic moment normal to the direction of magnetic field,

$W = MB (cos θ_{1} - cos θ_{2}) θ_{1} = 0^{\circ}, θ_{2} = 9 0^{\circ}] = 1.5 \times 0.22 (cos 0^{\circ} - cos 9 0^{\circ}) = 0.33 (1 - 0) = 0.33 J$

(ii) $θ_{1} = 0^{\circ}, θ_{2} = 18 0^{\circ}$ The work required to make the magnetic moment opposite to the direction of magnetic field,

$W = MB (cos θ_{1} - cos θ_{2}) = 1.5 \times 0.22 (cos 0^{\circ} - cos 18 0^{\circ}) = 0.33 [1 - (- 1)] = 0.66 J$

(b) For case (i) : $θ = θ_{2} = 9 0^{\circ}$

$∴ Torque, τ = MB sin θ = 1.5 \times 0.22 sin 9 0^{\circ} = 0.33 N - m$

For case (ii) : $θ = θ_{2} = 18 0^{\circ}$

$∴ Torque, τ = MB sin θ = MB sin 18 0^{\circ} = 0 N - m$

A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 1 0^{- 4} m^{2}$ , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 1 0^{- 2} T$ is set up at an angle of 30° with the axis of the solenoid?

Sol. $N = 2000, A = 1.6 \times 1 0^{- 4} m^{2}, I = 4 A$

(a) The magnetic moment along the axis of the solenoid,

$M = NIA = 2000 \times 4 \times 1.6 \times 1 0^{- 4} = 1.28 Am^{2}$

(b) The net force experienced by a magnetic dipole in a uniform magnetic field is zero. The torque $(τ)$ on the solenoid is given by,

$τ = MB sin θ = 1.28 \times 7.5 \times 1 0^{- 2} sin 3 0^{\circ} = 4.8 \times 1 0^{- 2} Nm$

The torque tends to align the axis of the solenoid (i.e., its magnetic moment vector $M$ ) parallel to the magnetic field $B$ .

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 lines (normal bisector) of the magnet. Sol. $M = 0.48 J / T, r = 10 cm = 0.1 m$ (a) The magnetic field at distance $r$ , from the centre of the magnet on the axis,

$∴ = B_{axis} = \frac{μ _{0}}{4 π} \frac{2 M}{r ^{3}} B_{axis} = \frac{4 π \times 1 0 ^{- 7} \times 2 \times 0.48}{4 π \times ( 0.1 ) ^{3}} 0.96 \times 1 0^{- 4} T = 0.96 G$

[The magnetic field is along the S-N direction] (b) The magnetic field at a distance of 10 cm on the equatorial line of the magnet

$B_{equatorial} = \frac{μ _{0} \times M}{4 π \times r ^{3}} = \frac{4 π \times 1 0 ^{- 7} \times 0.48}{4 π ( 0.1 ) ^{3}} = 0.48 G$

[The magnetic field is along the N-S direction]

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 of the magnet. 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 of the magnet at the same distance as the null-point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.) Sol. As the null point $N_{1}$ and $N_{2}$ lie on the axis (S-N) of the bar magnet, its south pole points towards the magnetic north pole of the earth as dip is zero. Clearly, for points $N_{1}$ and $N_{2}, B_{H}$ and $B_{axial}$ are in opposite direction and

$or B_{axial} = B_{H} \frac{μ _{o}}{4 π} \frac{2 M}{r ^{3}} = B_{H} \frac{μ _{o}}{4 π} \frac{2 M}{r ^{3}} = 0.36 G$

Magnetic field at a point P on the normal bisector (i.e., equatorial line) of the magnet at the same distance (r) from its centre O, i.e.,

$B_{equatorial} = \frac{μ _{0}}{4 π} \frac{M}{r ^{3}} = \frac{B _{H}}{2} B_{equatorial} = \frac{0.36 G}{2} = 0.18 G$

Since $B_{H}$ and $B_{equatorial}$ are in the same direction at point P, so total magnetic field at

$P, B = B_{H} + B_{equatorial} = 0.36 G + 0.18 G = 0.54 G .$

If the bar magnet in question 8 is turned round by $18 0^{\circ}$ , where will the new null points be located? Sol. When the magnet is turned around by 180°, its south pole will point toward magnetic south of the Earth [fig.] clearly, there are two null points ( $N_{1}$ and $N_{2}$ ) on normal bisector. The null point lie on the equatorial line of the magnet at a distance $r^{'}$ , from the centre O of the magnet, thus for points $N_{1}$ and $N_{2} -$ There are two null points on normal bisector $B_{H} = B_{equatorial}$ (in the plane of paper)

$B_{H} = \frac{μ _{o}}{4 π} \frac{M}{( r ^{'} ) ^{3}} \Rightarrow (r^{'})^{3} = \frac{μ _{o}}{4 π} \frac{M}{B _{H}} r^{'} = (\frac{μ _{o}}{4 π} \frac{M}{B _{H}})^{1/3},$

where $\frac{μ _{o}}{4 π} \frac{M}{B _{H}} = \frac{r ^{3}}{2}, r = 14 cm$ (from question 8)

$r^{'} = (\frac{r ^{3}}{2})^{1/3} = 14 \times 2^{- 1/3} = 11.1 cm$

from the centre of bar magnet.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for Class 12 Physics for all chapters, with exercise answers, important formulas, and clear solutions to help students understand concepts and practise questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 5

Comparison of Magnetic Materials: The solutions clearly compare diamagnetic, paramagnetic, and ferromagnetic materials, including their important properties and behaviour in a magnetic field.
Clear Explanation of Earth’s Magnetic Field: The solutions explain the magnetic meridian, geographic meridian, angle of dip, and horizontal component of the Earth’s magnetic field to help students understand related questions.
Bar Magnet and Solenoid: The solutions explain the similarity between a bar magnet and a current-carrying solenoid and help students understand the magnetic field produced by both.
Understanding Hysteresis: Retentivity and coercivity are explained step by step. The solutions also explain why soft iron is used for electromagnets and transformer cores, while materials with high retentivity are suitable for permanent magnets.
Clear Definitions and Units: Important quantities such as magnetic moment, magnetisation, magnetic intensity, susceptibility, and permeability are explained with their definitions and units to help students solve numerical questions correctly.

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NCERT Solutions Class 12 Physics Chapter 5 - Magnetism and Matter

1.0Class 12 Physics Chapter 5 Magnetism and Matter: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTION WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 5

Table of Contents

Frequently Asked Questions

Where can students access NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter?

What does NCERT Solutions Class 12 Physics Chapter 5 say about magnetic monopoles?

What should I know about the Earth’s magnetic field from NCERT Solutions Class 12 Physics Chapter 5?

Which quantities related to magnetism are defined in NCERT Solutions Class 12 Physics Chapter 5?

Why does hysteresis matter in NCERT Solutions Class 12 Physics Chapter 5?

Can NCERT Solutions Class 12 Physics Chapter 5 help with JEE and NEET preparation?

Can NCERT Solutions Class 12 Physics Chapter 5 help me choose between soft iron and steel for specific applications?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 5 - Magnetism and Matter

1.0Class 12 Physics Chapter 5 Magnetism and Matter: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTION WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 5

Table of Contents