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NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter

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

Class 12 Physics NCERT Solutions for Chapter 5 Magnetism and Matter by Vedantu explores the concepts of magnetism and its effects on matter. This chapter provides a comprehensive understanding of the magnetic properties of materials and the Earth's magnetism, which is important for applications in various technological and scientific fields. This chapter explains the classification of materials based on their magnetic properties, such as diamagnetic, paramagnetic, and ferromagnetic substances, each exhibiting unique behaviours in the presence of magnetic fields. This chapter provides a comprehensive understanding of magnetism and its applications in technology and daily life. With Vedantu's NCERT Solutions, you'll find step-by-step explanations of all the exercises in your textbook, ensuring you understand the concepts thoroughly.

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Table of Content
1. NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter - FREE PDF Download
2. Glance on Physics Chapter 5 Class 12 - Magnetism and Matter
3. Access NCERT Solutions for Class 12 Physics Chapter 5 – Magnetism and Matter
4. Overview of Deleted Syllabus for CBSE Class 12 Physics Magnetism and Matter
5. Conclusion
6. Other Study Material for CBSE Class 12 Physics Chapter 5
7. Chapter-Specific NCERT Solutions for Class 12 Physics
8. Related Links for NCERT Class 12 Physics in Hindi
9. Chapter-Specific NCERT Solutions for Class 12 Physics
FAQs


Glance on Physics Chapter 5 Class 12 - Magnetism and Matter

  • Chapter 5 of Class 12 Magnetism and Matter introduces the concept of magnetism and its effects on matter, exploring the magnetic properties of materials and the Earth's magnetism.

  • Diamagnetism, Paramagnetism, and Ferromagnetism are discussed, each with distinct characteristics and behaviours in magnetic fields. 

  • The relationship between Magnetic Field, Magnetization, and Magnetic Intensity is given.

  • Components of the Earth's magnetic field are discussed, including horizontal, vertical, and total intensity.

  • The chapter explains the uniform magnetic field inside a solenoid and the circular magnetic field in a toroid.

  • The properties and applications of permanent magnets and electromagnets are discussed.

  • This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 5 - Magnetism and Matter, which you can download as PDFs.

  • There are 7 fully solved questions in the exercise of class 12th Physics Chapter 5 Magnetism and Matter.

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Access NCERT Solutions for Class 12 Physics Chapter 5 – Magnetism and Matter

1. A short bar magnet placed with its axis at \[30{}^\circ \] 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?

Ans: Provided in the question,

Magnetic field strength \[B=0.25\,T\] 

Torque on the bar magnet, \[T=4.5\times {{10}^{-2}}J\] 

Angle between the given bar magnet and the external magnetic field, \[\theta =30{}^\circ \]

Torque is related to magnetic moment (M) as: 

\[T=MB\sin (\theta )\]

\[\Rightarrow M=\frac{4.5\times {{10}^{-2}}}{0.25\times \sin 30{}^\circ }=0.36J/T\]

Clearly, the magnetic moment of the magnet is \[0.36J/T\].


2. A short bar magnet of magnetic moment \[M=0.32\,J/T\] is placed in a uniform magnetic field of \[0.15\,T\]. If the bar is free to rotate in the plane of the field, which orientation and would correspond to its

a) Stable?

Ans: It is provided that moment of the bar magnet, \[M=0.32J/T\].

External magnetic field, \[B=0.15T\]

It is considered as being in stable equilibrium, when the bar magnet is aligned along the magnetic field. Therefore, the angle \[\theta \], between the bar magnet and the magnetic field is \[0{}^\circ \] .

Potential energy of the system \[=-MB\cos (\theta )\]

\[\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (0)=-4.8\times {{10}^{-2}}J\] 

Hence the potential energy is \[=-4.8\times {{10}^{-2}}J\]

b) Unstable equilibrium? What is the potential energy of the magnet in each case? 

Ans: It is provided that moment of the bar magnet, \[M=0.32J/T\]

External magnetic field, \[B=0.15T\]

When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium, \[\theta =180{}^\circ \]

Potential energy of the system is hence\[=-MB\cos (\theta )\]

\[\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (180{}^\circ )=4.8\times {{10}^{-2}}J\] 

Hence the potential energy is \[=4.8\times {{10}^{-2}}J\].


3. A closely wound solenoid of \[800\] turns and area of cross section \[2.5\times {{10}^{-4}}\,{{m}^{2}}\] carries a current of \[3.0A\]. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? 

Ans:  It is provided that number of turns in the solenoid, \[n=800\].

Area of cross-section, \[A=2.5\times {{10}^{-4}}{{m}^{2}}\]

Current in the solenoid, \[I=3.0A\]

A current-carrying solenoid is analogous to a bar magnet because a magnetic field develops along its axis, i.e., along its length joining the north and south poles.

The magnetic moment due to the given current-carrying solenoid is calculated as:

\[M=nIA=800\times 3\times 2.5\times {{10}^{-4}}=0.6J/T\]

Thus, the associated magnetic moment \[=0.6J/T\]


4. 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 \[30{}^\circ \] with the direction of applied field? 

Ans: Given is the magnetic field strength, \[B=0.25\,T\]

Magnetic moment, \[M=0.6\,/T\]

The angle, \[\theta \] between the axis of the turns of the solenoid and the direction of the external applied field is \[30{}^\circ \] .

Hence, the torque acting on the solenoid is given as: 

\[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =0.6\times 0.25\sin (30{}^\circ )\]

\[\Rightarrow \tau =7.5\times {{10}^{-2}}J\]

Hence the magnitude of torque is \[=7.5\times {{10}^{-2}}J\]


5. A bar magnet of magnetic moment \[1.5\,J/T\] lies aligned with the direction of a uniform magnetic field of \[0.22\,T\]. 

a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

Ans: Provided that,

Magnetic moment, \[M=1.5J/T\]

Magnetic field strength, \[B=0.22\,T\]

(i) Initial angle between the magnetic field and the axis is, \[{{\theta }_{1}}=0{}^\circ \]

Final angle between the magnetic field and the axis is, \[{{\theta }_{2}}=90{}^\circ \] 

The work that would be required to make the magnetic moment perpendicular to the direction of magnetic field would be:

\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]

\[\Rightarrow W=-1.5\times 0.22(\cos 90{}^\circ -\cos 0{}^\circ )\]

\[\Rightarrow W=-0.33(0-1)\]

\[\Rightarrow W=0.33\,J\]

(ii) Initial angle between the magnetic field and the axis, \[{{\theta }_{1}}=0{}^\circ \]

Final angle between the magnetic field and the axis, \[{{\theta }_{2}}=180{}^\circ \]

The work that would be required to make the magnetic moment opposite (180 degrees) to the direction of magnetic field is given as:

\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]

\[\Rightarrow W=-1.5\times 0.22(\cos 180{}^\circ -\cos 0{}^\circ )\]

\[\Rightarrow W=-0.33(-1-1)\]

\[\Rightarrow W=0.66J\]


b) What is the torque on the magnet in cases (i) and (ii)? 

Ans: For the first (i) case,

\[\theta ={{\theta }_{1}}=90{}^\circ \]

Hence the Torque, \[\vec{\tau }=\vec{M}\times \vec{B}\]

And its magnitude is:\[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =1.5\times 0.22\sin (90{}^\circ )\]

\[\Rightarrow \tau =0.33Nm\]

Hence the torque involved is \[=0.33Nm\]

For the second-(ii) case:

\[\theta ={{\theta }_{1}}=180{}^\circ \]

And its magnitude of the torque is:\[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =1.5\times 0.22\sin (180{}^\circ )\]

\[\Rightarrow \tau =0Nm\]

Hence the torque is zero.


6. 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 center allowing it to turn in a horizontal plane. 

a) What is the magnetic moment associated with the solenoid? 

Ans: Given is the number of turns on the solenoid, \[n=2000\]

Area of cross-section of the solenoid, \[A=1.6\times {{10}^{-4}}{{m}^{2}}\]

Current in the solenoid, \[I=4A\]

The magnetic moment inside the solenoid at the axis is calculated as:

\[M=nAI=2000\times 1.6\times {{10}^{-4}}\times 4=1.28A{{m}^{2}}\]


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 \[30{}^\circ \] with the axis of the solenoid?

Ans: Provided that,

Magnetic field, \[B=7.5\times {{10}^{-2}}T\]

Angle between the axis and the magnetic field of the solenoid, \[\theta =30{}^\circ \]

Torque, \[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =1.28\times 7.5\times {{10}^{-2}}\sin (30{}^\circ )\]

\[\Rightarrow \tau =4.8\times {{10}^{-2}}Nm\]

Given the magnetic field is uniform, and the force on the solenoid is zero. The torque on the solenoid is \[4.8\times {{10}^{-2}}Nm\].


7. A short bar magnet has a magnetic moment of \[0.48\,J/T\]. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of \[10\,cm\] from the center of the magnet on

a) the axis, 

Ans: Provided that the magnetic moment of the given bar magnet, \[M\]is \[0.48J/T\]

Given distance, \[d=10cm=0.1m\]

The magnetic field at d-distance, from the centre of the magnet on the axis is given by the relation:

\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}\]

here,

\[{{\mu }_{0}}=\] Permeability of free space\[=4\pi \times {{10}^{-7}}Tm/A\]

Substituting these values, \[B\] becomes as follows:

\[\Rightarrow B=\frac{4\pi \times {{10}^{-7}}}{4\pi }\frac{2\times 0.48}{{{0.1}^{3}}}\]

\[\Rightarrow B=0.96\times {{10}^{-4}}T=0.96\,G\]

The magnetic field is \[0.96G\] along the South-North direction.


b) the equatorial lines (normal bisector) of the magnet.

Ans: The magnetic field at a point which is \[d=10cm=0.1m\] away on the equatorial of the magnet is given as:

\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{d}^{3}}}\]

\[\Rightarrow B=\frac{4\pi \times {{10}^{-7}}}{4\pi }\frac{0.48}{{{0.1}^{3}}}\]

\[\Rightarrow B=0.48\times {{10}^{-4}}T=0.48G\]

The magnetic field is \[0.48G\]along the North-South direction.


Overview of Deleted Syllabus for CBSE Class 12 Physics Magnetism and Matter

Chapter

Dropped Topics

Magnetism and Matter

5.2.2 Bar Magnet as an Equivalent Solenoid (deleted only mathematical treatment)

5.2.3 The Dipole in a Uniform Magnetic Field (deleted only mathematical treatment)

Example 5.4

5.4 Earth’s Magnetism

5.41. Magnetic Declination and Dip

Table 5.2

5.6.2 Paramagnetism (deleted only Curie’s Law)

5.6.3 Ferromagnetism (deleted only Curie’s temperature; and Hysteresis)

5.7 Permanent Magnets and Electromagnets

Exercises 5.1, 5.2, 5.9–5.11, 5.13–5.25



Conclusion

NCERT Class 12 Physics Chapter 5 Exercise Solutions on Magnetism and Matter provided by Vedantu explains an in-depth exploration of magnetic properties and their practical applications. The Chapter provides a comprehensive understanding of how different materials interact with magnetic fields and the fundamental principles governing these interactions. By exploring topics such as magnetization, magnetic intensity, and hysteresis, as well as Earth's magnetism, this chapter equips students with essential knowledge for both academic and practical applications in technology and scientific research. From previous year's question papers, typically around 3–4 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills. 


Other Study Material for CBSE Class 12 Physics Chapter 5



Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.




Related Links for NCERT Class 12 Physics in Hindi

Discover relevant links for NCERT Class 12 Physics in Hindi, offering comprehensive study materials, solutions, and resources to enhance understanding and aid in exam preparation.




Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter

1. How many Topics are Present therein Magnetism and Matter?

Ans: Magnetism and matter include various important topics, which are very essential for the students of CBSE board. Maximum questions belong to this chapter; this chapter includes the Magnetic field, Diamagnetism, magnetic analogue, Ferromagnetism, and Electromagnets. It also includes Permanent Magnets, Magnetic Declination and Dip, Magnetism, and Gauss's law, The Earth magnetism, Magnetic intensity, and Various magnetic properties of materials. From this topic, various short types of questions, Long types of questions come in the examination.

2. What do you Mean by the Word Magnetism and Magnetic Fields?

Ans: Magnetism - It generally denotes the combined electromagnetic force. Magnetism is the force that arises between two magnets. It is caused because of the motion of electric charges. 


Magnetic Field - The magnetic field is a type of field where magnetic materials, electric charges and electric currents are moving around. So the field, where the particles get magnetically influenced is called the magnetic field. It is generally a vector field where the particles feel a force, perpendicular to its velocity.

3. How does the NCERT Solution of Magnetism and Matter help the Students to Solve Mathematical Problems?

Ans: The NCERT solution Class 12 Chapter 5 'Magnetism and matter' contains theory about magnetic fields, magnetic analogue, earth magnetism, magnetic properties of various materials GAUSS' law, etc. They are discussed briefly, and easily, so all the students can understand the theory. The theoretical knowledge helps the student to solve the mathematical problem and helps to reduce the fear of science, so they score very well in their examination.

4. If Students Have any Additional Doubts, How Does Vedantu Help the Student to Solve Their Queries?

Ans: If the students have additional queries regarding this chapter, they can easily clear their doubts by asking Vedantu's expert. There are also numerous questions present in the app, and by reading thoroughly they can get maximum understanding of the chapter. The experts are always present there to guide the students and help them to score more in the examination.


As the exams are not so far, students do not have a lot of time to prepare all the topics. So, without the best guidance, it is going to be very difficult. Vedantu is one of the best apps when it comes to guidance where every topic is discussed vastly and in-depth. They also help the students to prepare questions and answers, so that students can prepare themselves by attending online live classes. It is going to be a good decision to strengthen one’s understanding of lessons and gain knowledge rewarding question patterns.

5. What are the main topics covered in Chapter 5 - Magnetism And Matter?

Chapter 5 covers topics like magnetic poles and the concept of a magnetic field, magnetic field lines, Earth's magnetism, magnetic properties of materials, and the behaviour of magnetic materials.

6. How can I improve my understanding of the magnetic properties of materials discussed in this chapter?

To improve your understanding of magnetic properties, practice solving numerical problems and experiments related to magnetic materials. You can also conduct simple experiments at home to observe the behaviour of magnetic materials.

7. Is it essential to learn about Earth's magnetism in this chapter?

Yes, understanding Earth's magnetism is crucial in this chapter, as it helps you comprehend the Earth's magnetic field and its significance in navigation and other scientific applications.

8. Does magnetism in affect all matter?

Yes, magnetism affects all matter, but the extent of the effect varies. All materials respond to magnetic fields, though the nature and strength of their response differ. Materials can be classified as diamagnetic, paramagnetic, or ferromagnetic based on their magnetic properties. Diamagnetic materials are weakly repelled by magnetic fields, paramagnetic materials are weakly attracted, and ferromagnetic materials are strongly attracted and can retain their magnetization.

9. What is the cause of magnetism in matter, discussed in Magnetism and Matter NCERT solutions?

The cause of magnetism in matter is the motion of electric charges. At the atomic level, magnetism arises due to the movement of electrons around the nucleus and the spin of electrons on their axes. In ferromagnetic materials, magnetic domains (regions with aligned magnetic moments) contribute to a strong overall magnetic field when aligned.

10. What is the conclusion of Magnetism and Matter class 12 NCERT solutions?

The conclusion of studying Magnetism and Matter class 12 is that magnetic properties are intrinsic to materials and depend on the arrangement and motion of electrons. Understanding these properties helps in the development of various technologies, such as magnetic storage devices, electric motors, and medical imaging equipment. The chapter emphasizes the classification of materials based on their magnetic behaviour and their practical applications in everyday life.

11. What are the units mentioned in Magnetism and Matter NCERT solutions?

The primary units associated with magnetism are:

  • Magnetic Field (B): Tesla (T) or Gauss (G) (1 T = 10,000 G)

  • Magnetic Flux (Φ): Weber (Wb)

  • Magnetic Flux Density (B): Tesla (T)

  • Magnetic Permeability (μ): Henry per meter (H/m)

  • Magnetization (M): Ampere per meter (A/m)

  • Magnetic Field Intensity (H): Ampere per meter (A/m)

12. What is the theory of magnetism in ch 5 physics class 12 NCERT solutions?

The theory of magnetism in Magnetism and Matter NCERT  solutions explains how materials respond to magnetic fields due to the alignment of magnetic moments of electrons. The three main types of magnetic materials are:

  • Diamagnetic: Materials with no net magnetic moment; create an opposing magnetic field and are weakly repelled by magnets.

  • Paramagnetic: Materials with unpaired electrons that align with the magnetic field, causing weak attraction.

  • Ferromagnetic: Materials with magnetic domains that align strongly with an external magnetic field, resulting in strong attraction and permanent magnetization.

13. Is magnetism a form of matter?

As per Chapter 5 physics class 12 NCERT solutions, the magnetism is not a form of matter; it is a physical phenomenon that arises from the motion of electric charges. It is an interaction that affects materials, influencing their physical properties and behaviour in the presence of magnetic fields. Magnetism is a fundamental force, much like gravity and electromagnetism, that acts on particles with electric charge.