NCERT solutions for class 12 Physics chapter 5 Magnetism and Matter

NCERT Solutions for Class 12 Physics Chapter 5 - Magnetism and Matter

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The NCERT solutions class 12 physics chapter 5 Magnetism and Matter are created by subject matter experts to help the students with complete and clear solutions to all exercise questions of NCERT. Through NCERT solutions for Class 12 Magnetism and Matter, the students are capable of learning the most important questions, conceptual clarity and studying appropriately for the exams. These NCERT solutions for class 12 Physics Chapter 5 are also easy to reproduce as they are provided in step-by-step explanations and can be easily understood even on a complicated topic. Download these Magnetism and Matter solutions, without paying a dime and enrich your education now

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NCERT Solutions for Class 12 Physics Chapter 5: Download PDF
NCERT Solutions for Class 12 Physics Chapter 5: Exercises Question
NCERT Class 12 Physics Chapter 5 - Additional Questions
Approach to Solve Questions of Magnetism and Matter Class 12
What Extra Should Students Study Beyond the NCERT for JEE/NEET?
Important Topics from Class 12 Physics Chapter 5, Magnetism and Matter
Magnetism and Matter Class 12 NCERT Solutions: Concepts and Important Formulas
Importance Of NCERT Solutions For Class 12 Physics Chapter 5 Magnetism And Matter In Exams:
NCERT Solutions For Class 12 Physics Chapter-Wise

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NCERT Solutions for Class 12 Physics Chapter 5: Download PDF

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NCERT Solutions for Class 12 Physics Chapter 5: Exercises Question

5.1. 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 $4.5 \times 10^{- 2} J$ . What is the magnitude of magnetic moment of the magnet?

Answer:

Given,

The angle between axis of bar magnet and external magnetic field, θ = 30°

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, $τ$ = 4.5 x $10^{- 2}$ J

We know,

Torque experienced by a bar magnet placed in a uniform magnetic field is:

$\vec{τ} = \vec{m} \times \vec{B} \Rightarrow | \vec{τ} |$ = mBsin $θ$

$m = \frac{τ}{B s i n θ}$

$⟹ m = \frac{4.5 \times 10^{- 2} J}{0.25 T \times s i n 30^{\circ}}$

$∴$ m = 0.36 $J T^{- 1}$

Hence, the magnitude of the moment of the Bar magnet is 0.36 $J T^{- 1}$ .

5.2. A short bar magnet of magnetic moment $m = 0.32 J T^{- 1}$ 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?

Answer:

Given,

Magnetic moment of magnet, m = 0.32 $J T^{- 1}$

Magnetic field strength, B = 0.15 T

(a) Stable equilibrium: When the magnetic moment is along the magnetic field i.e. $θ = 0^{\circ}$

(b) Unstable equilibrium: When the magnetic moment is at 180° with the magnetic field i.e. $θ = 180^{\circ}$

U = $- \vec{m} . \vec{B}$ = -mBcos $θ$

By putting the given values:

U = (-0.32)(0.15)(cos $0^{\circ}$ ) = -0.048 J

Therefore, Potential energy of the system in stable equilibrium is -0.048 J

Similarly,

U = (-0.32)(0.15)(cos $180^{\circ}$ ) = 0.048 J

Therefore, Potential energy of the system in unstable equilibrium is 0.048J.

5.3 A closely wound solenoid of 800 turns and area of cross-section $2.5 \times 10^{- 4} m^{2}$ 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?

Answer:

In this case the magnetic field is generated along the axis / length of solenoid so it acts as a magnetic bar.

The magnetic moment is calculated as :-

$M = N I A$

or $= 800 \times 3 \times 2.5 \times 10^{- 4}$

or $= 0.6 J T^{- 1}$

5.4. 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?

Answer:

Given,

Magnetic field strength, B = 0.25 T

Magnetic moment, m = 0.6 JT −1

The angle between the axis of the solenoid and the direction of the applied field, $θ$ = 30°.

We know, the torque acting on the solenoid is:

$\vec{τ} = \vec{m} x \vec{B} \Rightarrow | \vec{τ} |$ = mBsin $θ$
$\Rightarrow | \vec{τ} |$ = (0.6 $J T^{- 1}$ )(0.25 T)(sin 30^o)= 0.075 J= 7.5 x $10^{- 2}$ J

The magnitude of torque is 7.5 x $10^{- 2}$ J.

5.5 (a) (i) A bar magnet of magnetic moment $1.5 J T^{- 1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?

Answer:

Given.

Magnetic moment, M= 1.5 $J T^{- 1}$

Magnetic field strength, B= 0.22 T

Now,

The initial angle between the axis and the magnetic field, $θ_{1}$ = 0°

Final angle, $θ_{2}$ = 90°

We know, The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

$W = - M B (c o s θ_{2} - c o s θ_{1})$

$⟹ W = - (1.5) (0.22) (c o s 90^{\circ} - c o s 0^{\circ}) = - 0.33 (0 - 1) = 0.33 J$

5.5 (a) (ii) A bar magnet of magnetic moment $1.5 T J T^{- 1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?

Answer:

The amount of work required for the given condition will be:-

$W = - M B [\cos Θ_{2} - \cos Θ_{1}]$

$W = - M B [\cos 180^{\circ} - \cos 0^{\circ}]$

or $W = 2 M B$

or $W = 2 \times 1.5 \times 0.22$

or $W = 0.66 J$

5.5 (b) A bar magnet of magnetic moment $1.5 J T^{- 1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the torque on the magnet in cases (i) and (ii)?

Answer:

For case (i):

$θ$ = $θ_{2}$ = 90°

We know, Torque, $τ = M B s i n θ$
$τ = (1.5) (0.22) s i n 90^{\circ} = 0.33 J$

For case (ii):

$θ$ = $θ_{2}$ = 180°

We know, Torque,

$τ = M B s i n θ$
$τ = (1.5) (0.22) s i n 180^{\circ} = 0$

5.6 (a) A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{- 4} m^{2}$ , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the magnetic moment associated with the solenoid?

Answer:

Given,

Number of turns, N = 2000

Area of the cross-section of the solenoid, $A = 1.6$ x $10^{- 4} m^{2}$

Current in the solenoid, I = 4 A

We know, The magnetic moment along the axis of the solenoid is:

m = NIA

m= (2000)(4 A)(1.6 x $10^{- 4} m^{2}$ = 1.28 $A m^{2}$

5.6 (b) A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{- 4} m^{2}$ , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{- 2} T$ is set up at an angle of 30° with the axis of the solenoid?

Answer:

Now,

Magnetic field strength, B = $7.5 \times 10^{- 2} T$

The angle between the magnetic field and the axis of the solenoid, $θ = 30^{\circ}$

Now, As the Magnetic field is uniform, the Force is zero

Also, we know,

$\vec{τ} = \vec{m} x \vec{B} \Rightarrow | \vec{τ} |$ = mBsin $θ$

$τ$ = (1.28 $J T^{- 1}$ )( $7.5 \times 10^{- 2} T$ )(sin $30^{\circ}$ )

$τ$ = 4.8 x $10^{- 2}$ J

Therefore, Force on the solenoid = 0 and torque on the solenoid = 4.8 x $10^{- 2}$ J

5.7 (a). A short bar magnet has a magnetic moment of $0.48 J T^{- 1}$ . 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 the axis,

Answer:

Given,

The magnetic moment of the bar magnet, m = 0.48 $J T^{- 1}$

Distance from the centre, d = 10 cm = 0.1 m

We know, The magnetic field at distance d, from the centre of the magnet on the axis is:

$B = \frac{μ_{0} m}{2 π r^{3}}$

$∴ B = \frac{4 π \times 10^{- 7} \times 0.48}{2 π (0.1)^{3}}$

$⟹$ $B =$ $0.96 \times 10^{- 4} T$

Therefore, the magnetic field on the axis, B = 0.96 G

Note: The magnetic field is along the S−N direction (like a dipole!).

5.7 (b). A short bar magnet has a magnetic moment of $0.48 J T^{- 1}$ 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 the equatorial lines (normal bisector) of the magnet.

Answer:

On the equatorial axis,

Distance,d = 10cm = 0.1 m

We know, the magnetic field due to a bar magnet along the equator is:

$B = - \frac{μ_{0} m}{4 π d^{3}}$

$∴ B = - \frac{4 π \times 10^{- 7} \times 0.48}{4 π (0.1)^{3}}$

$⟹$ B = $- 0.48 \times 10^{- 4} T$

Therefore, the magnetic field on the equatorial axis, B = 0.48 G

The negative sign implies that the magnetic field is along the N−S direction.

NCERT Class 12 Physics Chapter 5 - Additional Questions

1.(a) Answer the following questions regarding earth's magnetism: A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.

Answer:

The three independent quantities used to specify the earth's magnetic field are:

(f) The horizontal component of Earth's magnetic field ( $H_{E}$ ).

(ii) The magnetic declination (D): it is the angle between the geographic north and the magnetic north at a place.

(iii)The magnetic dip (1): It is the angle between the horizontal plane and the magnetic axis, as observed in the compass

1.(b) Answer the following questions regarding earth's magnetism. The angle of dip at a location in southern India is about $18^{\circ}$ . Would you expect a greater or smaller dip angle in Britain?

Answer:

We would expect a greater angle of dip in Britain. The angle of dip increases as the distance from equator increases.

(It is 0 at the equator and 90 degrees at the poles)

1.(c) Answer the following questions regarding earth's magnetism - If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

Answer:

The field lines go into the earth at the north magnetic pole and come out from the south magnetic pole and hence Australia being in the southern hemisphere. The magnetic field lines would come out of the ground at Melbourne.

1.(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

Answer:

The magnetic field is perpendicular at the poles and the magnetic needle of the compass tends to align with the magnetic field. Therefore the compass will get aligned in the vertical direction if is held vertically at the north pole.

1.(e) The earth's field, it is claimed, roughly approximates the field due to a dipole of magnetic moment $8 \times 10^{22} J T^{- 1}$ located at its centre. Check the order of magnitude of this number in some way

Answer:

Magnetic field

$B = \frac{μ_{μ} M}{4 π R^{3}}$

substituting the values

$\begin{aligned} R = 6.1 \times 10^{5} m \\ H = 4 π \times 10^{- T} \\ M = 8 \times 10^{22} J T^{- 1} \end{aligned}$

then

$B = 0.3 G$

1.(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?

Answer:

This may be possible due to the presence of minerals which are magnetic in nature.

2.(a) The Earth's magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

Answer:

Due to the constant but slow motion of the plates and change in the core, magnetic field due to Earth may change with time too. The time scale is in centuries for appreciable change.

2. (b) The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?

Answer:

The iron present in the core of the Earth is in the molten form. Hence it loses its ferromagnetism and is not regarded by geologists as a source of earth's magnetism.

2. (c) The charged currents in the outer conducting regions of the earth's core are thought to be responsible for Earth's magnetism. What might be the 'battery' (i.e., the source of energy) to sustain these currents?

Answer:

The radioactive materials might be the battery to sustain such currents.

2. (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth's field in such a distant past?

Answer:

The direction of the earth's magnetic field was recorded in rocks during solidification. By studying them, geologists can tell if the direction of the field had reversed.

2. (e) The earth's field departs from its dipole shape substantially at large distances (greater than about $30, 000 km$ ). What agencies may be responsible for this distortion?

Answer:

The earth's field departs from its dipole shape substantially at large distances (greater than about $30, 000 km$ ) due to the presence of ions in the ionosphere. These ions in motion generate magnetic field and hence distort the shape of a magnetic dipole.

2. (f) Interstellar space has an extremely weak magnetic field of the order of $10^{- 12}$ . Can such a weak field be of any significant consequence? Explain.

Answer:

This weak magnetic field can affect the motion of a charged particle in a circular motion. And a small deviation from its path in the vast interstellar space may have huge consequences.

3. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude $5.0 \times 10^{- 2}$ The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0 s^{- 1}$ . What is the moment of inertia of the coil about its axis of rotation?

Answer:

Given,

Number of turns, N = 16

Radius of the coil, r = 10 cm = 0.1 m

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 x $10^{- 2}$ T

Frequency of oscillations of the coil, f = 2.0 $s^{- 1}$

Now, Cross-section of the coil, A = $π r^{2}$ = $π \times (0.1)^{2} m^{2}$

We know that magnetic moment, m = NIA

m= (16)(0.75 A)( $π \times (0.1)^{2} m^{2}$ )

m= 0.377 $J T^{- 1}$

We know, the frequency of oscillation in a magnetic field is:

$f = \frac{1}{2 π} \sqrt{\frac{M B}{I}}$ (I = Moment of Inertia of the coil)

$⟹ I = \frac{M B}{4 π^{2} f^{2}}$

$⟹ I = \frac{0.377 \times 5 \times 10^{- 2}}{4 π^{2} 2^{2}}$

$I = 1.19 \times$ $10^{- 4}$ $k g m^{2}$

The moment of inertia of the coil about its axis of rotation is $1.19$ $\times$ $10^{- 4}$ $k g m^{2}$ .

4. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $22^{\circ}$ with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be $0.35 G$ . Determine the magnitude of the earth’s magnetic field at the place.

Answer:

Given,

The horizontal component of earth’s magnetic field, $B_{H}$ = 0.35 G

Angle made by the needle with the horizontal plane at the place = Angle of dip = $δ$ = $22^{\circ}$

We know, $B_{H}$ = B cos $δ$ , where B is earth's magnetic field

B = $B_{H}$ /cos $δ$ = 0.35/(cos $22^{\circ}$ ) = 0.377 G

The earth’s magnetic field strength at the place is 0.377 G.

5. At a certain location in Africa, a compass points $12^{\circ}$ west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points $60^{\circ}$ 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.

Answer:

Given,

The horizontal component of earth’s magnetic field, B H = 0.16 G

The angle of declination, $θ$ = $12^{\circ}$

The angle of dip, $δ$ = $60^{\circ}$

We know, $B_{H}$ = B cos $δ$ , where B is Earth's magnetic field

B = $B_{H}$ /cos $δ$ = 0.16/(cos $60^{\circ}$ ) = 0.32 G

Earth’s magnetic field is 0.32 G in magnitude lying in the vertical plane, $12^{\circ}$ west of the geographic meridian and $60^{\circ}$ above the horizontal.

6. 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.)

Answer:

Earth’s magnetic field at the given place, B = 0.36 G

The magnetic field at a distance d from the centre of the magnet on its axis is:

$B = μ_{0} m / 2 π d^{3}$

And the magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

$B^{'} = μ_{0} m / 4 π d^{' 3}$ = B/

( since d' = d, i.e same distance of null points.)

Hence the total magnetic field is B + B' = B + B/2 = (0.36 + 0.18) G = 0.54 G

Therefore, the magnetic field in the direction of earth’s magnetic field is 0.54 G.

7. If the bar magnet in Q.6 is turned around by $180^{\circ}$ , where will the new null points be located?

Answer:

Given, d = 14 cm

The magnetic field at a distance d from the centre of the magnet on its axis :

$B = μ_{0} m / 2 π d^{3}$

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial (perpendicular bisector) line.

The magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

$B = μ_{0} m / 4 π d^{' 3}$

Equating these two, we get:

$\frac{1}{2 d^{3}} = \frac{1}{4 d^{' 3}} ⟹ \frac{d^{' 3}}{d^{3}} = \frac{1}{2}$

d' = 14 x 0.794 = 11.1cm

The new null points will be at a distance of 11.1 cm on the normal bisector.

8 (a). A short bar magnet of magnetic movement $5.25 \times 10^{- 2} J T^{- 1}$ is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on its normal bisector?

Answer:

Given,

The magnetic moment of the bar magnet, $m = 5.25 \times$ $10^{- 2} J T^{- 1}$

The magnitude of earth’s magnetic field at a place, $H = 0.42 G = 0.42 \times$ $10^{- 4}$ T

The magnetic field at a distance R from the centre of the magnet on the normal bisector is:

$B = μ_{0} m / 4 π R^{3}$

When the resultant field is inclined at 45° with earth’s field, B = H

$B = H = 0.42 \times$ $10^{- 4}$ $T$

$R^{3} = μ_{0} m / 4 π B$ = $4 π \times 10^{- 7} \times 5.25 \times 10^{- 2} / 4 π \times 0.42 \times 10^{- 4}$

$R^{3} = 12.5 \times 10^{- 5} m^{3}$

Therefore, R = 0.05 m = 5 cm

8 (b). A short bar magnet of magnetic movement $5.25 \times 10^{- 2} J T^{- 1}$ is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on its axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Answer:

The magnetic field at a distance R from the centre of the magnet on its axis :

$B = μ_{0} m / 2 π R^{3}$

When the resultant field is inclined at 45° with earth’s field, B = H

$R^{3} = μ_{0} m / 2 π B$ = $4 π \times 10^{- 7} \times 5.25 \times 10^{- 2} / 2 π \times 0.42 \times 10^{- 4}$

$R^{3} = 25 \times 10^{- 5} m^{3}$

Therefore, R = 0.063 m = 6.3 cm

9. (a). Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?

Answer:

At high temperatures, alignment of dipoles gets disturbed due to the random thermal motion of molecules in a paramagnetic sample. But when cooled, this random thermal motion reduces. Hence, a paramagnetic sample displays greater magnetization when cooled.

9. (b). Why is diamagnetism, in contrast, almost independent of temperature?

Answer:

The magnetism in a diamagnetic substance is due to induced dipole moment. So the random thermal motion of the atoms does not affect it which is dependent on temperature. Hence diamagnetism is almost independent of temperature.

9. (c). If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

Answer:

A toroid using bismuth for its core will have slightly greater magnetic field than a toroid with an empty core because bismuth is a diamagnetic substance.

9. (d). Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

Answer:

We know that the permeability of ferromagnetic materials is inversely proportional to the applied magnetic field. Therefore it is more for a lower field.

9. (e). Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

Answer:

Since the permeability of ferromagnetic material is always greater than one, the magnetic field lines are always nearly normal to the surface of ferromagnetic materials at every point.

9. (f). Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Answer:

Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet for very strong magnetic fields.

10. (a). Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

Answer:

According to the graph between B (external magnetic field) and H (magnetic intensity) in ferromagnetic materials, magnetization persists even when the external field is removed. This shows the irreversibility of magnetization in a ferromagnet.

10 (b). The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?

Answer:

Material that has a greater area of hysteresis loop will dissipate more heat energy. Hence after going through repeated cycles of magnetization, a carbon steel piece dissipates greater heat energy than a soft iron piece, as the carbon steel piece has a greater hysteresis curve area.

10 (c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

Answer:

Ferromagnets have a record of memory of the magnetisation cycle. Hence it can be used to store memories.

11 (d). What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

Answer:

Ceramic, a ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer.

11 (e). A certain region of space is to be shielded from magnetic fields. Suggest a method.

Answer:

The region can be surrounded by a coil made of soft iron to shield from magnetic fields.

12. A long straight horizontal cable carries a current of 2.5 A in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer:

Given,

Current in the cable, I = 2.5 A

Earth’s magnetic field at the location, H = 0.33 G = 0.33 × 10^-4 T

The angle of dip, $δ$ = 0

Let the distance of the line of the neutral point from the horizontal cable = r

The magnetic field at the neutral point due to current carrying cable is:

$H_{n} = μ_{0} I / 2 π r$ ,

We know, Horizontal component of earth’s magnetic field, $H_{E}$ = $H c o s$ $δ$

Also, at neutral points, $H_{E} = H_{n}$

⇒ $H c o s$ $δ$ = $μ_{0} I / 2 π r$

$\Rightarrow 0.33 \times 10^{- 4} T \cos 0^{\circ} = \frac{4 π \times 10^{- 7 \times 2.5}}{2 π r}$

$\Rightarrow r = 1.515 c m$

Required distance is 1.515 cm.

13. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is $35^{\circ}$ . The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Answer:

Number of long straight horizontal wires = 4

The current carried by each wire = 1A

earth’s magnetic field at the place = 0.39 G

the angle of dip = 35^o

magnetic field due to infinite current-carrying straight wire

$B^{'} = \frac{μ_{0} I}{2 π r}$

r=4cm =0.04 m

$B^{'} = \frac{4 π \times 10^{- 7} \times 1}{2 π \times 4 \times 10^{- 2}}$

Magnetic field due to such 4 wires

$B = 4 \times \frac{4 π \times 10^{- 7} \times 1}{2 π \times 4 \times 10^{- 2}} = 2 \times 10^{- 5} T$

The horizontal component of the earth's magnetic field

$H = 0.39 \times 10^{- 4} c o s 35 = 0.319 \times 10^{- 4} T = 3.19 \times 10^{- 5} T$

The vertical component of the earth's magnetic field

$V = 0.39 \times 10^{- 4} s i n 35 = 0.22 \times 10^{- 4} T = 2.2 \times 10^{- 5} T$

At the point below the cable

$H^{'} = H - B = 3.19 \times 10^{- 5} - 2 \times 10^{- 5} = 1.19 \times 10^{- 5} T$

The resulting field is

$\sqrt{H^{' 2} + V^{2}} = \sqrt{(1.19 \times 10^{- 5})^{2} + (2.2 \times 10^{- 5})^{2}} = 2.5 \times 10^{- 5} T = 0.25 G$

14 (a). A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45 degree with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. Determine the horizontal component of the earth’s magnetic field at the location.

Answer:

Given,

Number of turns in the coil, n = 30

Radius of coil, r = 12cm = 0.12m

Current in the coil, I = 0.35A

The angle of dip, $δ$ = 45^o

We know, Magnetic fields due to current carrying coils, B = $μ_{0} n I / 2 r$

$B =$ $4 π \times 10^{- 7} \times 30 \times 0.35 / 2 \times 0.12$ = 5.49 $\times$ $10^{- 5}$ $T$

Now, Horizontal component of the earth’s magnetic field B_H= Bsin $δ$

$B_H= 5.49 \times10^{-5}T sin45^o$

$B_H=3.88\times10^{-5}T$ (Hint: Take sin45^o as 0.7)

$B_H=0.388 G$

14 (b). A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of $45^{\circ}$ with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Answer:

When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above, then the needle will reverse its direction. The new direction will be from east to west.

15. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^{\circ}$ , and one of the fields has a magnitude of $1.2 \times 10^{- 2}$ T. If the dipole comes to stable equilibrium at an angle of $15^{\circ}$ with this field, what is the magnitude of the other field?

Answer:

Given,

The magnitude of the first magnetic field, B₁ = 1.2 × 10^–2 T

The angle between the magnetic field directions, $θ$ = 60°

The angle between the dipole and the magnetic field $B_{1}$ is $θ_{1}$ = 15°

Let B 2 be the magnitude of the second magnetic field and M be the magnetic dipole moment

Therefore, the angle between the dipole and the magnetic field B₂ is $θ_{2}$ = $θ - θ_{1}$ = 45°

Now, at rotational equilibrium,

The torque due to field B₁ = Torque due to field B₂

$M B_{1} s i n θ_{1} = M B_{2} s i n θ_{2}$

$B_{2} = \frac{M B_{1} s i n θ_{1}}{M s i n θ_{2}} = \frac{1.2 \times 10^{- 2} \times s i n 15^{\circ}}{s i n 45^{\circ}}$

$B_{2} = 4.39 \times$ $10^{- 3}$ $T$

Hence the magnitude of the second magnetic field $= 4.39 \times$ $10^{- 3}$ $T$

16. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm $(m_{e} = 9.11 \times 10^{- 31} K g)$ . [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Answer:

The energy of electron beam = 18 eV

We can write:-

$E = \frac{1}{2} m v^{2}$

so $v = \sqrt{\frac{2 E}{M}}$

We are given horizontal magentic field : B = 0.40 G

Also, $B e v = \frac{m v^{2}}{r}$

We obtain, $r = \frac{1}{B e} \sqrt{2 E M}$

or $r = 11.3 m$

Using geometry, we can write:-

$\sin Θ = \frac{x}{r} = \frac{0.3}{11.3}$

and $y = r - r \cos Θ$

or $y = r (1 - \sqrt{1 - \sin^{2} Θ})$

or $y \approx 4 m m$

17. A sample of paramagnetic salt contains $2.0 \times 10^{24}$ atomic dipoles each of dipole moment $1.5 \times 10^{- 23} J T^{- 1}$ The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to $15 %$ . What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

Answer:

Given,

Magnetic field, $B_{1}$ = 0.64 T

Temperature, $θ_{1}$ = 4.2K

And, saturation = 15%

Hence, Effective dipole moment, $M_{1}$ = 15% of Total dipole moment

$M_{1}$ = 0.15 x (no. of atomic dipole × individual dipole moment)

$M_{1}$ = $0.15 \times 2 \times 10^{24} \times 1.5 \times 10^{- 23}$ = 4.5 $J T^{- 1}$

Now, Magnetic field, $B_{2}$ = 0.98 T and Temperature, $θ_{2}$ = 2.8 K

Let $M_{2}$ be the new dipole moment.

We know that according to Curie’s Law, $M \propto \frac{B}{θ}$

∴ The ratio of magnetic dipole moments

$\frac{M_{2}}{M_{1}} = \frac{B_{2} \times θ_{1}}{B_{1} \times θ_{2}}$

$⟹ M_{2} = \frac{B_{2} \times θ_{1}}{B_{1} \times θ_{1}} \times M_{1}$

$⟹ M_{2} = \frac{0.98 T \times 4.2 K}{0.64 T \times 2.8 K} \times 4.5 J T^{- 1}$ $= 10.336$ $J T^{- 1}$

Therefore, the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 $J T^{- 1}$

18. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Answer:

Given,

Radius of ring, r = 15cm = 0.15m

Number of turns in the ring, n = 3500

Relative permeability of the ferromagnetic core, $μ_{r}$ = 800

Current in the Rowland ring, I = 1.2A

We know,

Magnetic Field due to a circular coil, B = $\frac{μ_{r} μ_{0} n I}{2 π r}$

∴ B = $\frac{4 π \times 10^{- 7} \times 800 \times 3500 \times 1.2}{2 π \times 0.15}$ = 4.48T

Therefore, the magnetic field B in the core for a magnetising current is 4.48 T

19. The magnetic moment vectors $μ_{s}$ and $μ_{l}$ associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron, are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: $μ_{s} = - (e / m) S . μ_{l} = - (e / 2 m) l$ . Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

We know,

$μ_{l} = - (\frac{e}{2 m}) l$

$∴ μ_{l} = - (\frac{e}{2 m}) l$ is in expected from classical physics.

Now, the magnetic moment associated with the orbital motion of the electron is:

$μ_{l}$ = Current x Area covered by orbit = I x A= $(\frac{e}{T}) π r^{2}$

And, l = angular momentum = mvr = $m (\frac{2 π r}{T}) r$

(m is the mass of the electron having charge (-e), r is the radius of the orbit of by the electron around the nucleus and T is the time period.)

Dividing these two equations:

$\frac{μ_{l}}{l} = - \frac{e}{T} π r^{2} \times \frac{T}{m \times 2 π r^{2}} = - \frac{e}{2 m}$

$μ_{l} = (- \frac{e}{2 m}) l$ , which is the same result predicted by quantum theory.

The negative sign implies that $μ_{l}$ and l are anti-parallel.

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Approach to Solve Questions of Magnetism and Matter Class 12

Know All About the Material Magnetic Characteristics from Basic

The material behaves as a diamagnetic substance through weak repulsion against magnetic fields (bismuth and copper serve as examples).
Materials that exhibit paramagnetism show weak attractive forces (aluminium serves as an example material).
A ferromagnetic substance shows powerful magnetic attractions (including iron and cobalt).
The behaviour of materials under external magnetic fields (B) shows specific properties along with temperature-dependent responses defined through Curie’s Law.

$χ \propto \frac{1}{T}$

Magnetic Dipole in Uniform Magnetic Field

Use this to solve problems on:

Equilibrium position
Oscillation of magnetic dipole:

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$T = 2 π \sqrt{\frac{I}{M B}}$

where $I$ is moment of inertia of the dipole.

Understand Magnetic Susceptibility and Permeability

Relation:

$\begin{matrix} B = μ_{0} (H + M) = μ H \\ μ = μ_{0} (1 + χ) \end{matrix}$
Learn how $χ$ (susceptibility) and $μ$ (permeability) vary for different materials.

CBSE Class 12th Syllabus: Subjects & Chapters

Select your preferred subject to view the chapters

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

Concept Name	JEE	NCERT
Magnetic Field Lines	✅	✅
Bar Magnet As An Equivalent Solenoid	✅	✅
The Dipole In A Uniform Magnetic Field	✅	✅
Gauss Law Of Magnetism	✅	✅
Earth's Magnetic Field	✅	❌
Magnetisation And Magentic Intensity	✅	✅
Magnetic Properties Of Materials	✅	✅
Hysteresis Curve	✅	❌

JEE Main Highest Scoring Chapters & Topics

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Important Topics from Class 12 Physics Chapter 5, Magnetism and Matter

Some of the most important topics from Class 12 physics chapter 5, Magnetism and Matter, are mentioned here:

The Bar Magnet and Dipole

Magnetic Field Lines
Magnetic Dipole Moment
Bar Magnet as a Solenoid
Torque on a Bar Magnet in a Uniform Magnetic Field

Magnetism and Gauss's Law

Gauss's Law for Magnetism
Earth's Magnetism
Elements of Earth's Magnetic Field
Magnetic Declination (θ)
Magnetic Dip or Inclination (δ)
Horizontal Component (BH)

Magnetic Properties of Materials

Classification: diamagnetic, paramagnetic, ferromagnetic substances
Hysteresis Loop

Magnetism and Matter Class 12 NCERT Solutions: Concepts and Important Formulas

1. Magnetic Dipole and Magnetic Moment

A magnetic dipole consists of two equal and opposite magnetic poles separated by a distance.
Magnetic moment $(M) =$ Pole strength $(m) \times$ Separation (d)

$M = m \times d$

The SI unit of the magnetic moment is $A \cdot m^{2}$ .

2. Torque on a Magnetic Dipole in a Uniform Magnetic Field

A magnetic dipole experiences a torque when placed in a uniform magnetic field.

$τ = M B \sin θ$

where $τ$ is torque, $M$ is the magnetic moment, $B$ is the magnetic field, and $θ$ is the angle between M and B .

3. Bar Magnet as an Equivalent Solenoid

A bar magnet produces a similar field pattern as a solenoid.
Magnetic field at the axial position of a bar magnet:

$B_{axial} = \frac{μ_{0}}{4 π} \cdot \frac{2 M}{r^{3}}$

Magnetic field at the equatorial position:

$B_{equatorial} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$

4. Earth's Magnetism

Earth behaves like a huge bar magnet with its magnetic field approximately aligned along the geographic axis.
Important elements of Earth's magnetic field:
Magnetic Declination (D): Angle between geographic and magnetic meridian.
Magnetic Inclination (I): Angle between Earth's magnetic field and the horizontal plane.
Horizontal Component of Earth's Magnetic Field (B_H):

$B_{H} = B \cos I$

5. Gauss's Law for Magnetism

The net magnetic flux through any closed surface is always zero, meaning magnetic monopoles do not exist.

$\oint B \cdot d A = 0$

6. Magnetic Properties of Materials

Diamagnetic Materials: Weakly repelled by a magnetic field (e.g., copper, water).
Paramagnetic Materials: Weakly attracted to a magnetic field (e.g., aluminium, oxygen).
Ferromagnetic Materials: Strongly attracted and retained magnetism (e.g., iron, cobalt, nickel).

7. Curie's Law for Paramagnetism

Magnetic susceptibility $(x)$ is inversely proportional to temperature ( T ):

$χ = \frac{C}{T}$

where $C$ is Curie's constant

Importance Of NCERT Solutions For Class 12 Physics Chapter 5 Magnetism And Matter In Exams:

Every year in the CBSE board Physics paper, up to 5 marks questions are asked from the class 12 chapter 5 physics. Solutions of NCERT Class 12 Physics is important for competitive exams like NEET and JEE Main also. If you combine chapters 4 and 5, maybe around 2-4 questions can come in the NEET exam and 2-3 questions can come in JEE mains exam. From the NCERT Class 12 Physics Chapter 5, you can get these marks in your pocket easily.

NCERT Solutions For Class 12 Physics Chapter-Wise

There are a total 14 chapters present in class 12 physics, Chapter-wise magnetism and matter class 12 solutions are listed below:

NCERT solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT solutions for Class 12 Physics Chapter 4 Moving charges and magnetism

NCERT solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

NCERT solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT solutions for Class 12 Physics Chapter 11 Dual nature of radiation and matter

NCERT solutions for Class 12 Physics Chapter 12 Atoms

NCERT solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

Also, check NCERT Books and NCERT Syllabus here:

NCERT solutions subject-wise

Also, check NCERT Exemplar Class 12 Solutions

Frequently Asked Questions (FAQs)

Q: What is magnetism?

Magnetism is a force caused by moving electric charges or inherent properties of materials leading to attraction or repulsion. Earth itself acts as a magnet.

Q: Why does a suspended magnet point north-south?

it aligns with Earth’s magnetic field making one end the north pole and the other the south pole.

Q: What are diamagnetic, paramagnetic, and ferromagnetic materials?

Diamagnetic: Weakly repelled (e.g. - copper).

Paramagnetic: Weakly attracted but temporary (e.g. - aluminium).

Ferromagnetic: Strongly attracted and retains magnetism (e.g. - iron).

Q: Why is Earth a magnet?

Electric currents in its molten core create a magnetic field, guiding compasses and protecting us from solar winds.

Q: What happens if you cut a magnet?

Each piece forms a new magnet with both north and south poles—magnetic monopoles do not exist.

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Can i improve my 12 result this year in PCM

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If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

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Rahul Mandal

8 Sep'25

how may marks should i get in my 12th board exams to get fee concesion in sastra university

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SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

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After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

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4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

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CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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12 arts previous years board paper 2025

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 12 Physics Chapter 5 - Magnetism and Matter

NCERT Solutions for Class 12 Physics Chapter 5: Download PDF

NCERT Solutions for Class 12 Physics Chapter 5: Exercises Question

NCERT Class 12 Physics Chapter 5 - Additional Questions

Related eBooks & Sample Papers

Approach to Solve Questions of Magnetism and Matter Class 12

Know All About the Material Magnetic Characteristics from Basic

Magnetic Dipole in Uniform Magnetic Field

Aakash Repeater Courses

Understand Magnetic Susceptibility and Permeability

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

Important Topics from Class 12 Physics Chapter 5, Magnetism and Matter

Magnetism and Matter Class 12 NCERT Solutions: Concepts and Important Formulas

1. Magnetic Dipole and Magnetic Moment

2. Torque on a Magnetic Dipole in a Uniform Magnetic Field

3. Bar Magnet as an Equivalent Solenoid

4. Earth's Magnetism

BH=Bcos⁡I5. Gauss's Law for Magnetism

6. Magnetic Properties of Materials

7. Curie's Law for Paramagnetism

Importance Of NCERT Solutions For Class 12 Physics Chapter 5 Magnetism And Matter In Exams:

NCERT Solutions For Class 12 Physics Chapter-Wise

Also Read

Also, check NCERT Books and NCERT Syllabus here:

NCERT solutions subject-wise

Also, check NCERT Exemplar Class 12 Solutions

Questions related to CBSE Class 12th

Popular CBSE Class 12th Questions

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$B_{H} = B \cos I$

5. Gauss's Law for Magnetism