Pearson | PTE
Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally
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.
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.
Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally
Free download magnetism and matter ncert solutions PDF for CBSE exam.
Answer:
The three independent quantities used to specify the earth’s magnetic field are:
(i) The horizontal component of Earth's magnetic field ( ).
(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
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
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
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.
Magnetic field
substituting the values
then
5.1 (f) Answer the following questions regarding earth’s magnetism
This may be possible due to the presence of minerals which are magnetic in nature.
5.2 (a) Answer the following questions
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 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 (c) Answer the following questions
The radioactive materials might be the battery to sustain such currents.
5.2 (d) Answer the following questions
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) 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
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.
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 J
We know,
Torque experienced by a bar magnet placed in a uniform magnetic field is:
T = m x B = mBsin
m = 0.36
Hence, the magnitude of the moment of the Bar magnet is 0.36 .
Given,
Magnetic moment of magnet, m = 0.32
Magnetic field strength, B = 0.15 T
(a) Stable equilibrium: When the magnetic moment is along the magnetic field i.e.
(b) Unstable equilibrium: When the magnetic moment is at 180° with the magnetic field i.e.
(c) We know that,
U = - m.B = -mBcos
By putting the given values:
U = (-0.32)(0.15)(cos ) = -0.048 J
Therefore, Potential energy of the system in stable equilibrium is -0.048 J
Similarly,
U = (-0.32)(0.15)(cos ) = 0.048 J
Therefore, Potential energy of the system in unstable equilibrium is 0.048J.
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 :-
or
or
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, = 30°.
We know, the torque acting on the solenoid is:
= m x B = mBsinθ
= (0.6 )(0.25 T)(sin 30 o )
= 0.075 J
= 7.5 x J
The magnitude of torque is 7.5 x J.
Answer:
Given.
Magnetic moment, M= 1.5
Magnetic field strength, B= 0.22 T
Now,
The initial angle between the axis and the magnetic field, = 0°
Final angle, = 90°
We know, The work required to make the magnetic moment normal to the direction of the magnetic field is given as:
= 0.33 J
5.7 (a)
The amount of work required for the given condition will be:-
or
or
or
For case (i):
= = 90°
We know, Torque,
= 0.33 J
For case (ii):
= = 180°
We know, Torque,
= 0
Given,
Number of turns, N = 2000
Area of the cross-section of the solenoid, x
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 )
= 1.28
Now,
Magnetic field strength, B =
The angle between the magnetic field and the axis of the solenoid,
Now, As the Magnetic field is uniform, the Force is zero
Also, we know,
= mxB = mBsinθ
= (1.28 )( )(sin )
= 4.8 x J
Therefore, Force on the solenoid = 0 and torque on the solenoid = 4.8 x J
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 T
Frequency of oscillations of the coil, f = 2.0
Now, Cross-section of the coil, A = =
We know, Magnetic moment, m = NIA
= (16)(0.75 A)( )
= 0.377
We know, frequency of oscillation in a magnetic field is:
(I = Moment of Inertia of the coil)
The moment of inertia of the coil about its axis of rotation is .
Given,
The horizontal component of earth’s magnetic field, = 0.35 G
Angle made by the needle with the horizontal plane at the place = Angle of dip = =
We know, = B cos , where B is earth's magnetic field
B = /cos = 0.35/(cos ) = 0.377 G
The earth’s magnetic field strength at the place is 0.377 G.
Given,
The horizontal component of earth’s magnetic field, B H = 0.16 G
The angle of declination, =
The angle of dip, =
We know, = B cos , where B is Earth's magnetic field
B = /cos = 0.16/(cos ) = 0.32 G
Earth’s magnetic field is 0.32 G in magnitude lying in the vertical plane, west of the geographic meridian and above the horizontal.
Given,
The magnetic moment of the bar magnet, m = 0.48
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:
Therefore, the magnetic field on the axis, B = 0.96 G
Note: The magnetic field is along the S−N direction (like a dipole!).
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 =
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.
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:
And the magnetic field at a distance d' from the centre of the magnet on the normal bisector is:
= 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 , where will the new null points be located?
Given, d = 14 cm
The magnetic field at a distance d from the centre of the magnet on its axis :
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:
Equating these two, we get:
d' = 14 x 0.794 = 11.1cm
The new null points will be at a distance of 11.1 cm on the normal bisector.
Given,
The magnetic moment of the bar magnet,
The magnitude of earth’s magnetic field at a place, T
The magnetic field at a distance R from the centre of the magnet on the normal bisector is:
When the resultant field is inclined at 45° with earth’s field, B = H
=
Therefore, R = 0.05 m = 5 cm
The magnetic field at a distance R from the centre of the magnet on its axis :
When the resultant field is inclined at 45° with earth’s field, B = H
= =
Therefore, R = 0.063 m = 6.3 cm
5.16 Answer the following questions
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?
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
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.
We know that the permeability of ferromagnetic materials is inversely proportional to the applied magnetic field. Therefore it is more for a lower field.
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.
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.
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.
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.
Ferromagnets have a record of memory of the magnetisation cycle. Hence it can be used to store memories.
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.
The region can be surrounded by a coil made of soft iron to shield from magnetic fields.
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, = 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:
,
We know, Horizontal component of earth’s magnetic field, =
Also, at neutral points,
⇒ =
Required distance is 1.515 cm.
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
r=4cm =0.04 m
magnetic field due to such 4 wires
The horizontal component of the earth's magnetic field
the horizontal component of the earth's magnetic field
At the point below the cable
The resulting field is
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, = 45 o
We know, Magnetic fields due to current carrying coils, B =
Now, Horizontal component of the earth’s magnetic field = Bsin
(Hint: Take sin45 o as 0.7)
When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of in the anticlockwise sense looking from above, then the needle will reverse its direction. The new direction will be from east to west.
Given,
The magnitude of the first magnetic field, B 1 = 1.2 × 10 –2 T
The angle between the magnetic field directions, = 60°
The angle between the dipole and the magnetic field is = 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 = = 45°
Now, at rotational equilibrium,
The torque due to field B 1 = Torque due to field B 2
Hence the magnitude of the second magnetic field
The energy of electron beam = 18 eV
We can write:-
so
We are given horizontal magentic field : B = 0.40 G
Also,
We obtain,
or
Using geometry, we can write:-
and
or
or
Given,
Magnetic field, = 0.64 T
Temperature, = 4.2K
And, saturation = 15%
Hence, Effective dipole moment, = 15% of Total dipole moment
= 0.15 x (no. of atomic dipole × individual dipole moment)
= = 4.5
Now, Magnetic field, = 0.98 T and Temperature, = 2.8 K
Let be the new dipole moment.
We know that according to Curie’s Law,
∴ The ratio of magnetic dipole moments
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
Given,
Radius of ring, r = 15cm = 0.15m
Number of turns in the ring, n = 3500
Relative permeability of the ferromagnetic core, = 800
Current in the Rowland ring, I = 1.2A
We know,
Magnetic Field due to a circular coil, B =
∴ B = = 4.48T
Therefore, the magnetic field B in the core for a magnetising current is 4.48 T
We know,
is in expected from classical physics.
Now, the magnetic moment associated with the orbital motion of the electron is:
= Current x Area covered by orbit = I x A
=
And, l = angular momentum = mvr
=
(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:
, which is the same result predicted by quantum theory.
The negative sign implies that 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:
Important formulas of ncert solutions class 12 magnetism and matter are given below:
Magnetic Field due to a Straight Current-Carrying Conductor
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.
Terms | Magnetic | Electrostatic |
Dipole moment | p | m |
Equitorial field(short dipole) | -p/4πϵ0r3 | -μ0m/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.
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-
Comprehensive Coverage: The magnetism class 12 ncert solutions encompass all the important topics and concepts in the Class 12 Physics chapter "Magnetism and Matter."
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.
Detailed Explanations: class 12 magnetism and matter ncert solutions come with detailed explanations, helping students grasp the underlying principles and solving techniques.
Clarity and Simplicity: The magnetism and matter class 12 ncert solutions are presented in clear and simple language, making complex concepts more accessible.
Practice and Self-Assessment: These questions and answers provide ample opportunities for students to practice and evaluate their knowledge and problem-solving skills.
Exam Preparation: They are valuable resources for preparing for board exams and other competitive exams like JEE and NEET.
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.
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.
The heading covered in the chapter 5 of Class 12 NCERT Physics book are:
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.
Application Date:21 November,2024 - 20 December,2024
Application Date:21 November,2024 - 20 December,2024
Application Date:02 December,2024 - 16 December,2024
Application Date:02 December,2024 - 16 December,2024
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
Re-evaluate Your Study Strategies:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.
I hope this information helps you.
Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
Type C: For candidates scoring between 40% and 50%.
Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.
Thank you and wishing you all the best for your bright future.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters