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NCERT Class 12 Physics Chapter 4 Notes Moving Charges and Magnetism - Download PDF

NCERT Class 12 Physics Chapter 4 Notes Moving Charges and Magnetism - Download PDF

Edited By Vishal kumar | Updated on Feb 02, 2024 04:59 PM IST

Welcome to our updated NCERT Moving Charges and Magnetism class 12 notes, carefully crafted by the Careers360 experts. We've covered all of the important concepts and formulas in these CBSE class 12 physics ch 4 notes in an easy-to-understand manner. They are not only informative; they are also intended to improve your understanding and performance in exams, assignments, and homework.

We understand how important well-organized notes are when preparing for board exams or difficult entrance exams like those for medical and engineering courses. Our goal is to make your learning relevant and effective. Also, these class 12 physics chapter 4 notes are available in PDF format which gives access to free download anytime anywhere.

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Force Applied To A Moving Charge:

A moving charge is the source of the magnetic field.

Assume q is a positive charge moving in a uniform magnetic field Band velocity v

1643017914598 [ k is constant]

Where in S.I. system k=1

F=qvBsinθ and

\vec{F}=q(\vec{v}\times \vec{B})

Magnetic field strength B:

In the equation, we can observe this F=qvBsinθ, if





then F=B

As a result, magnetic field strength can be defined as the force experienced by a unit charge moving at unit velocity perpendicular to the magnetic field's direction.

Some cases for this:

  1. If θ=0o or 180o


There will be no force acting on a charged particle moving parallel to the magnetic field.

  1. When

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A charged particle in a magnetic field will not experience any force when it is at rest.

  1. When 1643017919054 then the force will be maximum


The highest force will be experienced by a charged particle moving perpendicular to the magnetic field.

  • S.I. unit of magnetic field intensity:

The S.I unit is tesla (T).




That is 1643017916232

Then B=1T

When a charge of 1C at a velocity of 1 m/s encounters a force of 1N while moving at a right angle to a magnetic field, the strength of the magnetic field is called 1T.

  • Biot-Savart’s Law:

Because of the current element dl, the strength of magnetic flux density or magnetic field at a point P (dB) will be dependent on


After we combine them:



1643017917953 , 1643017923161


dB will be perpendicular to the plane containing dl

  • Biot-law Savart's applications:

A magnetic field (B) is maintained at the centre of a circular current-carrying coil with a radius of r.

The magnetic field in the centre of a circular coil with n turns will be, if there are n turns.


The number of turns of the coil is denoted by n. I will be the coil's current, and r will be the radius of the coil.

Magnetic field caused by a straight conductor carrying current:


The perpendicular distance of the conductor from the place where the field is to the measured value will be denoted by a.

Φ1 and Φ2 will be the angles formed by the conductor's two ends meeting at the location.



A semi-circular current-carrying conductor will have a magnetic field at its centre.


The magnetic field at the centre of a circular current-carrying conductor arc with an angle of θ at the centre will be,


Ampere Circuital Law:

In a vacuum, the magnetic field line integral around any closed passage is mu zero times the entire current via the closed path. that is

\int B.dl=\mu_0I

  • 7. Application of Ampere’s circuital law:

  • Magnetic field caused by a current-carrying solenoid, 1643017929592


  • Magnetic field caused by a toroid or endless solenoid,


  • Charged Particle Motion in a Uniform Electric Field: –

The path of a charged particle in an electric field is called a parabola.

Equation of the parabola be


Where x is the electric field's width.

y is the particle's deviation from its straight path.

v is the charged particle's speed. q is the particle's charge.

E denotes the strength of the electric field. Let m be the particle's mass.

In a uniform magnetic field B, the path of a charged particle moving at 1643017916471 producing an angle 1643017922938 with B will be a helix. Because the charged particle will not be supplied a force by the component of velocity 1643017937569, the particle will move forward with a fixed velocity in the direction of B. The other component, 1643017926040, will produce the force 1643017924917, which will provide the necessary centripetal force to the charged particle moving along a circular route with radius r.

Centripetal force =



Angular velocity of rotation =


Frequency of rotation =


The time period of revolution =


  • Cyclotron:

This is a technology that we use to accelerate positively charged particles and thereby energize them. This can be achieved by immersing the particle in a perpendicular magnetic field that oscillates and an electric field that oscillates. A circular path will be followed by the particle.

Centripetal force=magnetic Lorentz force


the radius of the circular path

Time for travelling a semicircular path =


=constant When 1643017908318 is the particle's highest velocity and 1643017930907 is the maximum radius of its path, we can say that


The maximum kinetic energy of the particle =


The time period of the oscillating electric field =


Cyclotron frequency =


Cyclotron angular frequency =


  • Force acting on a current-carrying conductor kept in a magnetic field will be


I be the current flowing through the conductor in this case.

B be the strength of the magnetic field.

l denotes the conductor's length.

1643017915554 be the angle formed by the current direction and the magnetic field.

If 1643017916951

There will be no force acting on a conductor if it is kept parallel to the magnetic field.



F will be maximum


The conductor will be experiencing maximal force if it is kept normal to the magnetic field.

  • The force between two parallel current-carrying conductors: –

  1. If the current flows in the same direction, the two conductors will be drawn together by a force.

  2. When the current is flowing in the opposite direction, the two conductors repel each other with equal force.

  • Torque Experienced on a Current-Carrying Coil Kept in a Magnetic Field:

\tau=\vec{M}\times \vec{B}



where M be the magnetic dipole moment of the coil

Where n is the coil's number of turns.

I represent the current flowing through the coil.

B is the magnetic field's intensity.

The coil's area is denoted by the letter A.

The angle between the magnetic field B and the coil's normal to the plane will be α

  • Moving Coil Galvanometer:

This is based on the premise that if a coil carrying electricity is held in a magnetic field, it will experience torque. Because of the phosphor bronze strip, there is a restoring torque, which returns the coil to its usual position.

In equilibrium,

Deflecting torque = Restoring torque


where Galvanometer constant



The deflection made if the unit current is passed through the galvanometer is the current sensitivity of the galvanometer.


The deflection caused by a unit potential difference placed across the galvanometer is known as voltage sensitivity.


  • The maximum sensitivity of the galvanometer is having some conditions: -

If a modest current causes a big deflection, the galvanometer is said to be sensitive.


  • Galvanometer To Voltmeter And Ammeter Conversion:

  1. By connecting a high resistance to a galvanometer, it can be turned to a voltmeter.

Total resistance of voltmeter = Rg + R

Where Rg be the galvanometer resistance.

R be the resistance added in series.

Current through the galvanometer =


Here V is the potential difference across the voltmeter.


By connecting a low resistance in parallel with a galvanometer, it can be transformed into an ammeter (shunt)
Shunt =


where Rg be the galvanometer’s resistance.

The effective resistance of the ammeter will be,


Significance of NCERT Class 12 Physics Chapter 4 Notes

Comprehensive Revision: These Moving Charges and Magnetism class 12 notes provide a concise summary of Chapter 4 from the NCERT Class 12 Physics textbook, allowing students to revise key concepts before exams.

Understanding Core Concepts: Class 12 physics chapter 4 notes help students understand important topics in moving charges and magnetism, such as magnetic fields, the magnetic force on a current-carrying conductor, and the Biot-Savart law.

Preparation for Competitive Exams: Because the cbse class 12 physics ch 4 notes cover key topics from the CBSE physics syllabus, they are useful resources for competitive exams such as VITEEE, BITSAT, JEE Main, NEET, and others. They ensure that students are adequately prepared for these exams by emphasising fundamental principles.

Offline Study: The availability of the Moving Charges and Magnetism notes class 12 in PDF format enables students to study offline, allowing them to access the material at any time and from any location without the need for an internet connection.

Structured Learning Aid: These ch 4 physics class 12 notes, organised into points or sections, provide a structured approach to studying the chapter, allowing students to focus on specific topics sequentially and understand them more effectively.

Quick Reference: They are used as a quick reference guide for students who need to review specific concepts or formulas while solving problems or preparing for exams, saving time and effort.

Overall, these Physics class 12 chapter 4 notes pdf help students learn, revise, and prepare for exams by ensuring a solid understanding of moving charges and magnetism concepts.

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Frequently Asked Question (FAQs)

1. What are the Different Applications of Ampere's Circuital Law?

The Class 12 notes on moving charges and magnetism contain a detailed description of the various applications of Ampere's Circuital Law. Examples include (1) a long current-carrying wire producing magnetism, (2) a long current-transmitting cylinder producing a magnetic field, and (3) a current-carrying hollow cylinder producing a magnetic field. The derivations for all of these applications are provided in the Chapter 4 Moving Charges and Magnetism notes for Class 12 Physics.

2. What do you mean by Cyclotron?

As mentioned in Moving Charges and Magnetism Notes, the cyclotron was one of the first particle accelerators. Despite a flurry of developments since then, prototypes are still utilised in the early phases of specialised multi-stage particle accelerators. According to ncert notes for Class 12 Physics chapter 4, the attribute of a magnetic force's impact on a moving charge is also used to bend the latter along a semi-circular trajectory.

3. What's the Connection Between Magnetism and Moving Charges according to CBSE Class 12 Physics chapter 4 notes?

Moving charges or charge flow cause magnetism. Magnetic fields also exert forces on charge flow, which in turn exerts forces on other magnets.. Because of the presence of continuous moving charges, such a phenomenon occurs.

Because electricity is a flow of moving charge, the link between electricity and magnetism includes both attraction and repulsion between distinct charged particles, as well as force exertion inside such charges. Electromagnetism is the word for the interaction of electricity and magnetism, as seen in the Class 12 Moving charges and magnetism notes

4. What are the Biot Savart Law's Applications?

The Biot Savart Law is discussed in the ncert notes for Class 12 Physics chapter 4. The law's main applications are: (1) calculating magnetic responses at the atomic and molecular level, and (2) determining velocity in aerodynamics theory.

5. What do you understand about the Right-Hand Rule according to chapter 4 of Class 12 Physics?

The Right-Hand Rule is used to determine the direction of a (+ive) travelling charge's magnetic force. In this scenario, we stretch our right hand's thumb and first two fingers to point to the conductor's motion, the magnetic field, and the induced current, with the thumb pointing to the conductor's motion, the first finger to the magnetic field, and the middle finger to the induced current. It's critical to understand the theory, definition, and explanation of the formula for numerical, as well as how to use it.


Get answers from students and experts

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)


Option 2)

\; K\;

Option 3)


Option 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)


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)


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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