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NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

Edited By Vishal kumar | Updated on Sep 09, 2023 10:26 AM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 5 – Access and Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter: Having the magnetism class 12 ncert solutions is like having a helpful tool when you're solving homework problems and assignments. These class 12 magnetism and matter ncert solutions cover a total of twenty-five questions, starting from 5.1 to 5.15 in the exercise section and from 5.16 to 5.25 in the additional exercise section. They are created by experts who explain things in a simple way, making it easy for students to understand each step.

The NCERT solutions are very important for CBSE board exam. There are also a few derivations mentioned in chapter 5 Class 12 Physics. The comparisons of electric dipole in chapter 1 and magnetic dipole in Physics chapter 5 Class 12 will help in the NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter. Some of the comparisons based on NCERT are given in NCERT Class 12 Physics Chapter 5 pdf.

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You have studied in the first chapter that an isolated charge can be obtained. Can you obtain a magnetic monopole? If we cut a magnet into several parts each part will act as a magnet with south and north pole or magnetic dipole. This shows that magnetic monopole cannot exist. And you will study such properties of the magnet in Magnetism and Matter. Questions related to all the topics of Chapter 5 Physics Class 12 are covered in Class 12 Physics Chapter 5 NCERT solutions. While studying magnetism Class 12 you can compare it with electrostatics.

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NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

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NCERT solutions for class 12 physics chapter 5 magnetism and matter: Exercises Question and Answer

5.1(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:

(i) 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 (I): It is the angle between the horizontal plane and the magnetic axis, as observed in the compass

5.1(b) Answer the following questions regarding earth’s magnetism. The angle of dip at a location in southern India is about 18°. 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)

5.1 (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.

5.1 (d) Answer the following questions regarding earth’s magnetism

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.

5.2 (a) Answer the following questions

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.


5.2 (b) Answer the following questions

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 it ferromagnetism and not regarded by geologists as a source of earth's magnetism.

5.2 (d) Answer the following questions

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 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.

5.2 (e) Answer the following questions

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.

5.2 (f) Answer the following questions

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.

5.3. 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:

T = m x B = mBsin \theta

m = \frac{T}{Bsin\theta}

\implies m = \frac{4.5\times10^{-2}J}{0.25T\times sin30^{\circ}}

\therefore m = 0.36 JT^{-1}

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

5.4. A short bar magnet of magnetic moment m = 0.32 JT ^ { -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 JT^{-1}

Magnetic field strength, B = 0.15 T

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

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

(c) We know that,

U = - m.B = -mBcos \theta

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.

5.5 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\ =\ NIA

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

or =\ 0.6\ JT^{-1}

5.6. If the solenoid in Exercise 5.5 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, \theta = 30°.

We know, the torque acting on the solenoid is:

\tau = m x B = mBsinθ
= (0.6 JT^{-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.

5.7 a) 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: (i) 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, \theta_1 = 0°

Final angle, \theta_2 = 90°

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

W = - MB ( cos \theta_2 - cos \theta_1)

\implies W = - (1.5)(0.22) ( cos 90^{\circ} - cos 0^{\circ}) = -0.33(0-1)
= 0.33 J

5.7 (b) A bar magnet of magnetic moment 1.5 JT^{-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):

\theta = \theta_2 = 90°

We know, Torque, \tau = MBsin\theta
= (1.5)(0.22)sin90^{\circ}
= 0.33 J

For case (ii):

\theta = \theta_2 = 180°

We know, Torque,

\tau = MBsin\theta
= (1.5)(0.22)sin180^{\circ}
= 0

5.8 (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

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

= 1.28 Am^2

5.8 (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.

(b) 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, \theta = 30^{\circ}

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

Also, we know,

\tau = mxB = mBsinθ

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

= 4.8 x 10^{-2} J

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

5.9. 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 = \pi r^2 = \pi \times (0.1)^2 m^2

We know, Magnetic moment, m = NIA

= (16)(0.75 A)( \pi \times (0.1)^2 m^2 )

= 0.377 JT^{-1}

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

f = \frac{1}{2\pi}\sqrt{\frac{MB}{I}} (I = Moment of Inertia of the coil)

\implies I = \frac{MB}{4\pi^2f^2}

\implies I = \frac{0.377\times5\times10^{-2}}{4\pi^22^2}

I = 1.19\times 10^{-4} kgm^2

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

5.10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22 \degree 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 = \delta = 22 \degree

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

B = B_{H} /cos \delta = 0.35/(cos 22 \degree ) = 0.377 G

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

5.11. At a certain location in Africa, a compass points 12 \degree 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 \degree 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, \theta = 12^{\circ}

The angle of dip, \delta = 60^{\circ}

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

B = B_{H} /cos \delta = 0.16/(cos 60 \degree ) = 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.

5.12 (a). A short bar magnet has a magnetic moment of 0.48 JT^{-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 JT^{-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{\mu_{0} m}{2\pi r^3}

\therefore B= \frac{4\pi\times10^{-7}\times 0.48}{2\pi (0.1)^3}

\implies 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. 12 (b). A short bar magnet has a magnetic moment of 0.48 JT^{-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{\mu_{0} m}{4\pi d^3}

\therefore B=- \frac{4\pi\times10^{-7}\times 0.48}{4\pi (0.1)^3}

\implies 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.

5.13). 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 = \mu_{0}m/2\pi d^{3}

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

B' = \mu_{0}m/4\pi d'^3

= B/2 ( 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.

5.14. If the bar magnet in exercise 5.13 is turned around by 180 \degree , 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 = \mu_{0}m/2\pi 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 = \mu_{0}m/4\pi d'^3

Equating these two, we get:

\\ \frac{1}{2d^3} = \frac{1}{4d\ '^3}\\ \\ \implies \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.

5.15 (a). A short bar magnet of magnetic movement 5.25 \times 10 ^{-2} JT^{-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}\ JT^{-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 = \mu_{0}m/4\pi 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 = \mu_{0}m/4\pi B = 4\pi\times10^{-7}\times5.25\times10^{-2}/4\pi\times0.42\times10^{-4}

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

Therefore, R = 0.05 m = 5 cm

Magnetism and matter class 12 solutions: Additional Exercise

5.16 Answer the following questions

(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.

5.16 Answer the following questions

(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.

5.16 Answer the following questions

(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.

5. 16 (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.

5. 16 (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.

5.16 (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.

5.17 (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.

5.17 (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.

5.17 (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.

5.17 (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.

5.17(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.

5.18. A long straight horizontal cable carries a current of 2.5 A in the direction 10 \degree south of west to 10 \degree north of east. The magnetic meridian of the place happens to be 10 \degree 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, \delta = 0

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

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

H_{n} = \mu_{0}I/2\pi r ,

We know, Horizontal component of earth’s magnetic field, H_{E} = Hcos \delta

Also, at neutral points, H_{E}= H_n

Hcos \delta = \mu_{0}I/2\pi r

1643697701109

\Rightarrow r = 1.515 cm

Required distance is 1.515 cm.

5.19 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 \degree . 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 0

magnetic field due to infinite current-carrying straight wire

B'=\frac{\mu_0I}{2\pi r}

r=4cm =0.04 m

B'=\frac{4\pi\times10^{-7}\times1}{2\pi \times4\times10^{-2}}

magnetic field due to such 4 wires

B=4\times \frac{4\pi\times10^{-7}\times1}{2\pi \times4\times10^{-2}}=2\times10^{-5}T

The horizontal component of the earth's magnetic field

H=0.39\times10^{-4}cos35=0.319\times10^{-4}T=3.19\times10^{-5}T

the horizontal component of the earth's magnetic field

V=0.39\times10^{-4}sin35=0.22\times10^{-4}T=2.2\times10^{-5}T

At the point below the cable

H'=H-B=3.19\times10^{-5}-2\times10^{-5}=1.19\times10^{-5}T

The resulting field is

\sqrt{H'^2+V^2}=\sqrt{(1.19\times10^{-5})^2+(2.2\times10^{-5})^2}=2.5\times10^{-5}T=0.25G

5.20(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, \delta = 45 o

We know, Magnetic fields due to current carrying coils, B = \mu_{0}nI/2r

B = 4\pi \times10^{-7}\times30\times0.35/2\times0.12

= 5.49 \times 10^{-5} T

Now, Horizontal component of the earth’s magnetic field = Bsin \delta

= 5.49 \times 10^{-5} T � sin45^o

= 3.88\times 10^{-5} T (Hint: Take sin45 o as 0.7)

=0.388 G

5.21. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60 \degree , 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 \degree with this field, what is the magnitude of the other field?

Answer:

Given,

The magnitude of the first magnetic field, B 1 = 1.2 × 10 –2 T

The angle between the magnetic field directions, \theta = 60°

The angle between the dipole and the magnetic field B_{1} is \theta_{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 B 2 is \theta_{2} = \theta - \theta_{1} = 45°

Now, at rotational equilibrium,

The torque due to field B 1 = Torque due to field B 2

MB_{1}sin\theta_{1}= MB_{2}sin\theta_{2}

B_{2} = \frac{MB_{1}sin\theta_{1}}{Msin\theta_{2}} = \frac{1.2\times10^{-2}\times sin15\degree}{sin45\degree}

= 4.39 \times 10^{-3} T

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

5.23. A sample of paramagnetic salt contains 2.0 \times 10 ^{24} atomic dipoles each of dipole moment 1.5 \times 10 ^{-23} JT^{-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, \theta_{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 JT^{-1}

Now, Magnetic field, B_{2} = 0.98 T and Temperature, \theta_{2} = 2.8 K

Let M_{2} be the new dipole moment.

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

∴ The ratio of magnetic dipole moments

\\\frac{M_{2}}{M_{1}}= \frac{B_{2}\times \theta_{1}}{B_{1}\times \theta_{2}} \\ \implies M_{2}= \frac{B_{2}\times \theta_{1}}{B_{1}\times \theta_{1}}\times M_{1} \\\implies M_{2}= \frac{0.98T\times4.2K}{0.64T\times2.8K}\times 4.5 JT^{-1} = 10.336 JT^{-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 JT^{-1}

5.24. 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, \mu_{r} = 800

Current in the Rowland ring, I = 1.2A

We know,

Magnetic Field due to a circular coil, B = \frac{\mu_{r} \mu_{0}nI}{2\pi r}

∴ B = \frac{4\pi\times10^{-7}\times800\times3500\times1.2}{2\pi \times0.15} = 4.48T

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

5.25) The magnetic moment vectors \mu _s and \mu _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:
\mu _ s = - ( e / m ) S\\\\.\: \: \: \: \: \: \mu _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,

\mu_{l} = -(\frac{e}{2m})l

\therefore \mu_{l} = -(\frac{e}{2m})l is in expected from classical physics.

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

\mu_{l} = Current x Area covered by orbit = I x A

= (\frac{e}{T})\pi r^2

And, l = angular momentum = mvr

= m(\frac{2\pi 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{\mu_{l}}{l} = -\frac{e}{T}\pi r^2\times \frac{T}{m\times2\pi r^2} = -\frac{e}{2m}

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

The negative sign implies that \mu_{l} and l are anti-parallel.

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:

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

Important formulas of ncert solutions class 12 magnetism and matter are given below:

  • Magnetic Field due to a Straight Current-Carrying Conductor

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B = (μ₀ * I) / (2π * r)

Where: B is the magnetic field, μ₀ (mu-zero) is the permeability of free space, I is the current, and r is the distance from the conductor.

  • Magnetic Field due to a Circular Current Loop:

B = (μ₀ * I) / (2R)

Where: B is the magnetic field, μ₀ is the permeability of free space, I is the current, and R is the radius of the loop.

  • Magnetic Field due to a Solenoid:

B = μ₀ * n * I

Where: B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.

  • Magnetic Permeability (μ):

μ = (1 + χ) * μ₀

Where: μ is the permeability of the material, χ is the susceptibility, and μ₀ is the permeability of free space.

  • Magnetic Dipole Moment (μ):

μ = I * A

Where μ is the magnetic dipole moment, I is the current, and A is the area of the loop.

The comparisons of an electric dipole and magnetic dipole:

Terms
Magnetic
Electrostatic
Dipole moment
p
m
Equitorial field(short dipole)
-p/4πϵ0r3
0m/4πr3
Axial field(short dipole)
2p/4πϵ0r3
μ02m/4πr3
Torque in an external field
vector product of p and E
vector product of m and B
Energy(external field)
-p.E(dot product)
-m.B(dot product)
  • Gauss's law in magnetism is another important topic discussed in Physics Class 12 chapter 5.

  • Earth's magnetism- This section of NCERT Class 12 talks about earth magnetism and terms like declination, dip etc.

  • Topic on magnetisation and magnetising intensity, susceptibility etc are detailed in Class 12 NCERT Physics chapter 5 and the questions related to this are discussed in the NCERT Solutions for Class 12 Physics Chapter 5.

  • Magnetisation curve, magnetic properties of ferromagnetic, diamagnetic and paramagnetic materials and a comparison of permanent magnet and electromagnets are discussed in the chapter Magnetism and Matter.

Magnetism and Matter Class 12: Important Topics

The NCERT Chapter 5 Physics Class 12 Magnetism and Matter discuss bar magnets, magnetic materials and earth magnetism. To understand the NCERT solutions for Class 12 Physics Chapter 5, the following main topics are to be referred to with the help of the Class 12 NCERT book.

  • The bar magnet-This topic of ch5 Physics Class 12 give ideas on bar magnet, field due to bar magnet and give explanation and proof to show that bar magnet is an equivalent solenoid.

  • Dipole in a magnetic field- The next topic of chapter 5 Physics Class 12 is the dipole in a magnetic field, the time period of oscillations of the dipole, the potential energy of magnetic dipole and comparison of an electrostatic and magnetic dipole. Questions based on this are explained in magnetic properties of matter Class 12 solutions. The analogy is given below-

Key Features of magnetism and matter questions and answers

  1. Comprehensive Coverage: The magnetism class 12 ncert solutions encompass all the important topics and concepts in the Class 12 Physics chapter "Magnetism and Matter."

  2. Varied Difficulty Levels: Physics class 12 chapter 5 Questions range from basic to advanced, allowing students to practice and assess their understanding at different levels.

  3. Detailed Explanations: class 12 magnetism and matter ncert solutions come with detailed explanations, helping students grasp the underlying principles and solving techniques.

  4. Clarity and Simplicity: The magnetism and matter class 12 ncert solutions are presented in clear and simple language, making complex concepts more accessible.

  5. Practice and Self-Assessment: These questions and answers provide ample opportunities for students to practice and evaluate their knowledge and problem-solving skills.

  6. Exam Preparation: They are valuable resources for preparing for board exams and other competitive exams like JEE and NEET.

  7. Foundation for Advanced Study: The concepts covered serve as the foundation for more advanced topics in physics and related fields.

These features make ncert solutions class 12 magnetism and matter an essential resource for students, aiding them in their studies and exam preparation.

NCERT Exemplar Class 12 Solutions

Also Check NCERT Books and NCERT Syllabus here:

Subject Wise Solutions

Importance of NCERT solutions for class 12 physics chapter 5 magnetism and matter in exams:

In last year CBSE board Physics paper, 3 marks questions were asked from the class 12 chapter 5 physics ncert solutions. Solutions of NCERT Class 12 Physics is important for competitive exams like NEET and JEE Main also. If you combine chapters 3 and 4, 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.

Frequently Asked Question (FAQs)

1. What are the main headings covered in the NCERT syllabus of Magnetism and Matter?

The heading covered in the chapter 5 of Class 12 NCERT Physics book are:

  • The bar-magnet
  • Magnetic field lines
  • Bar magnet as equivalent solenoid
  • Magnetic dipole
  • Electrostatic analog
  • Magnetism and Gauss's laws
  • Earth's magnetism
  • Magnetisation and magnetic intensity
  • Magnetic materials
  • Hysteresis curve
  • Electromagnet and Permanent magnet
2. Which material should I refer for more problems on CBSE Chapter Magnetism and Matter?

NCERT exemplar for Class 12 Physics provides problems on Magnetism and Matter. Also refer CBSE board previous year papers for a better score in the board exam.

3. Are class 12 physics chapter 5 ncert solutions helpful in JEE Main exam preparation?

NCERT solutions for class 12 physics chapter 5 are useful resources for preparing for the JEE Main exam, as they provide detailed explanations of the concepts and principles covered in the syllabus.

The Class 12 Physics Chapter 5, "magnetism and matter," covers a range of topics that are important for the JEE Main exam. These include the nature of magnetism and the properties of magnets, the behaviour of magnetic materials, the motion of charged particles in a magnetic field, the electron theory of magnetism, and the Zeeman effect.

Studying the NCERT solutions for these topics can help candidates gain a better understanding of the concepts and develop the problem-solving skills necessary to do well on the JEE Main exam. 

4. What are the properties of magnets?

Magnets have several properties, including a north pole and a south pole, and the ability to attract or repel other magnets or certain types of metal. They can also be magnetized or demagnetized, and they can be affected by the presence of other magnets or electric fields.

5. What is the electron theory of magnetism?

The electron theory of magnetism is a model that explains the origin of magnetism in terms of the movement of electrons in an atom. According to this theory, the electrons in an atom are responsible for the magnetic properties of a material, and the direction and strength of the magnetism depend on the arrangement of the electrons.

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Have a question related to CBSE Class 12th ?

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  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

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  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
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  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

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hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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