CBSE Class 12 Physics Revision Notes Chapter 4 Moving  Charges and  Magnetism

The Class 12 Physics Chapter 4 is  ‘Moving  Charges and  Magnetism’. From this chapter, students will be able to learn all concepts about charges, current induction, magnetic properties, and the effect of magnetism on charged particles like electrons, protons, and some wire carrying current. 

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Extramarks Class 12 Physics Chapter 4 Notes cover all the study material, resolving students’ doubts and  strengthening their fundamental concepts with  exercises, solutions and derivations as last-minute revision notes to maximise their potential and help them achieve their goal.. 

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Key Topics Covered In Class 12 Physics Chapter 4 Notes 

Class 12 Physics Chapter 4 Notes explains  topics on magnetism and electric charges. It illustrates with examples electrostatic field lines, steady currents- Is Newton’s third law valid here?, the Lorentz force, Ampere’s circuital law and the Biot-savart law, its similarity between Gauss’s law and Coulomb’s law.  . The end-text questions include exercises and additional exercises to check your understanding of the chapter.   Class 12 Extramarks’ Notes cover all the key topics which students might find challenging while studying. Therefore, these notes are prepared in such a way that they can get to the point answers without wasting much time on a single subject.

Following are the topics included in Extramarks Notes of Class 12 Physics Chapter 4:

Magnetic Field:

Every magnet produces a magnetic field at all surrounding points. A fixed charge has an electric field in the surroundings. Similarly, a moving charge creates a magnetic field that exerts a force on the moving charge. 

Some important Properties:

• It is a scalar quantity.
• The magnetic field is represented by B.
• The SI unit is Tesla/T or W/m2 weber per metre square.
• The CGS unit of the magnetic field is Gauss or fostered.

Effect of a charged particle in a Magnetic field:

A Force ‘F’ acts on the particle when any charge ‘q’ is placed in the magnetic field ‘B’, with a velocity ‘v’. The force is given by ‘F’ = qvBsin?. Where ? is the angle between velocity and magnetic field, the kinetic energy of particles does not change due to magnetic force as the magnetic force is perpendicular to the velocity. 

• Case 1:
• When the value of ? is 0o or 180o
• When the value of ? is 0o or 180o, the charged particle moves parallel to the magnetic field. And hence it continues to move in a straight line with constant velocity.

• Case 2:
• When the value of ? is 90o.
• If the value of ? is 90o, the charged particle moves perpendicular to the magnetic field. Hence the path traced by charge is a circle.

The force on charge by the magnetic field is given as,

F = qvB sin90 

F = qvB…….(1)

F = mv 2 R……..(2)

From (1) and (2), 

R = mvqB

Angular velocity= ⍵ = vR = qBm

Time period for revolution = T = 2π⍵ = 2πmqB 

Frequency = 1T= qB2πm

To learn more about the aforementioned cases, students can sign up  at Extramarks’ website and get access to our Class 12 Physics Chapter 4 Notes and also, check out a repository of study material they might need to supplement their studies to avoid last minute hassle. It’s better to plan, prepare and stay ahead with no regrets.

Magnetic Force:

Magnetic Force is defined as the force of attraction or repulsion exerted between the pole of any magnet and electrically charged moving particles. It is denoted by F.

Any charge q placed in an electric field experiences a force F =qE

F = qQr(4?εo) r2, where r is the unit vector along with r.

Lorentz Force:

As explained in our Class 12 Physics Chapter 4 Notes, Lorentz Force is the force exerted on a charged particle by magnetic and electric fields simultaneously. 

Lorentz Force F is given by,

F = FE + FB

F = electric field force + magnetic field force

F = qE + qVB

Where F = Lorentz force

q = charge on particle

E = electric field

B = magnetic field

V = velocity of the particle

Special Case:

• When the charged particle is in a vacuum.
• When in a vacuum, the particle does not undergo frequent collisions, and hence the particle moves in the direction perpendicular to the magnetic field B. As a result, it moves in a circular path with a radius r = mv/qB.
• If the electric force to the charged particle is equal and opposite to the magnetic force, then the net force on a charged particle is zero.

Therefore, v = E/B.

The motion of charge in a magnetic field:

When any charged particle moves in a magnetic field, the magnetic force becomes perpendicular to the particle’s velocity. Hence no work is done, and there is no change in the magnitude of velocity. 

The Force F = qvB acts as a centripetal force and produces a circular motion of charge perpendicular to the magnetic field. This motion in a circular field creates a spiral shape. But the electric field in the y-direction imparts acceleration along that axis. Due to this, the particle acquires velocity in the y-direction and starts moving in a spiral motion. All the parameters remain the same as that of circular motion.

R = ν / αB

T = 2π / αB

ν = αB / 2π

ω = αB

If a velocity component is a present parallel to a magnetic field, it makes the particle move along both fields. And the motion becomes a winding one. 

Some important Properties as covered in our Class 12 Physics Chapter 4 Notes:

The presence of two fields, magnetic field and electric field, gives the following uses:

• Charge moves in the electric and magnetic fields.
• J.J. Thomson experiment. ( Where a specific charge of an electron is measured).
• Cyclotron ( The charged particles are accelerated).

The magnetic field from Current element/ Biot-Savart Law:

A magnetic field is produced from the motion of electrical charges. Biot-Savart law gives the relation between current and magnetic field. The law relates magnetic field to magnitude, direction, length and proximity of electric current. 

According to Biot-Savart law, the magnetic field produced by a current-carrying conductor, having length dl, current I flowing through the conductor, at a point from distance r is given as:

dB = µo / 4 π I dl * r/r3 or

dB = µo / 4 π I dl sin?/r2

Where ? is the current direction angle and µo is the absolute permeability of free space. 

 If you have already been through the NCERT book and are still unable to comprehend or solve exercises given in the chapter, you may  refer  to our Class 12 Physics Chapter 4 Notes, you need to sign up at Extramarks’ website.   

Some important Properties:

• The SI unit is Wm-2.
• The CGS unit is Gauss.
• 1 gauss = 10-4, Tesla.
• The direction of magnetic field dB is the same as that of I dl *r.

Ampere’s Circuital Law:

Ampere’s circuital law is the current distribution method that helps to calculate the magnetic field in any closed area or vacuum. Biot-Savart law and Ampere’s circuital law do the same job. But the latter method uses a high case symmetry. 

Ampere’s circuital law states that the line of integral magnetic field or B, calculated around any closed path in a vacuum, is 110 times the total current flowing through it. 

B dl = µo I

Here, B = magnetic field

Dl = length of the current-carrying conductor,

I = current 

µo = absolute permeability of free space. 

This law holds good for any closed path of any size and shape. The point for consideration can be anywhere as the relation is independent of distance from the conductor. 

Ampere’s circuital law has been further explained with  more examples and exercise questions and solutions  in our Class 12 Physics Chapter 4 Notes. 

Solenoid:

A solenoid comprises insulated, long thin copper wire closely wrapped  around a helix shape which  produces a  magnetic field  when the current flows through it.

The magnetic field at any point inside the  solenoid is given as,

B = µo nI

Where B = magnetic field inside solenoid

N = number of turns of wire

I = current flowing through solenoid

The right-hand thumb rule gives the direction of this magnetic field. It is commonly used to obtain a uniform magnetic field. 

Long Solenoid:

As explained in our Class 12 Physics Chapter 4 Notes,  long  solenoid has  more considerable length as compared to the radius. It consists of a long wire wrapped  multiple times around a helix.. Each turn on the solenoid is regarded as a circular loop. The magnetic field on the long solenoid is the vector sum of all magnetic fields from all coil turns.

The magnetic field inside a solenoid is,

B = µo nI

Where B = magnetic field inside solenoid

N = number of turns of wire

I = current flowing through solenoid

The magnetic field at a point on one end of the solenoid is given as,

B =(µo nI/2)

Toroid:

Toroid consists of an insulated wire wound around a doughnut-shaped iron form. A solenoid bent into the close circular loop is a toroid. A toroid has no endpoints. Hence the loss of magnetic flux is minimised. Flux linkage is maximum as compared to the solenoid. 

These are some of the standard applications of toroid:

• Power transformers.
• Low-level inductors.
• Power inductors.
• Low-level transformers.
• Current transformers.

The magnetic field inside a toroid is given as, 

B =(µo nI/2πr)

Where I = current flowing through toroid

r = average radius of the toroid

N = number of turns per unit length 

Torque:

The twisting force that tends to cause the rotation of objects is called torque. The axis of rotation is the point of the object where it rotates. 

Some important Properties:

• Torque is denoted by ?.
• ? = F x r,

Where F = force applied, r = distance between the centre of the axis of rotation and the point where pressure is applied 

Torque on current loop:

When a steady current is passed through a rectangular coil, placed in a uniform magnetic field, it experiences a torque. The loop does not experience any force. This behaviour is similar to an electric dipole in a uniform electric field. 

Construction: 

A rectangular loop PQRS is placed in a uniform magnetic field, surrounding bar magnets on the left-hand and right-hand sides. There are two brushes attached to a circuit. Due to the presence of magnets, the rectangular coil undergoes anticlockwise direction due to torque developed. 

• Case 1:

A rectangular loop PQRS is placed in a uniform magnetic field B. There is no force exerted on the arms of loop PS and QR. These are perpendicular to the arm PQ and exert a force F1 directed into the plane of the loop. The magnitude of the force can be stated as,

F1 = IzB

Similarly, the arm RS experience a force F2 directed out of the plane,

F2 = IzB = F1

As both the forces F1 and F2 are equal in magnitude and opposite in direction, there is no net force acting on the loop. Hence, torque is formed on the loop. The torque developed tends to rotate the coil in the anticlockwise direction. 

? = F1(y/2) + F2(y/2)

= IzB(y.2) + IzB(y/2)

= I(y x z)B

= IAB

Where I = current flowing through the coil

B = magnetic field

A = area of rectangular coil = length x breadth = y x z

• Case 2: 

If the rectangular coil is placed not along the magnetic field, it is in at some angle. Consider θ to be the angle between the magnetic field and the normal to the rectangular coil. 

The forces acting on the arms of the coil, QR and SP, are equal and opposite in direction. They operate along the coil axis, which connects the centre of mass of arms QR and SP. The collinear forces along the axis cancel out each other and result in no net force or torque. 

Forces on arm PQ and RS are F1 and F2, respectively. These are also equal and opposite to each other. 

F1 = F2 = IzB

Since these two are not collinear, they result in a couple. However, the effect of torque is less than in Case 1, when the loop was into a magnetic field. The magnitude of the torque on the loop can be stated as,

? = F1(y/2) sinθ + F2(y/2) sinθ

= I(y x z) B sinθ

= IAB sinθ

From the two equations obtained from two cases, ? = IAB and ? = IAB sinθ, the torque can be expressed as the vector product of the magnetic moment of the coil and the magnetic field. Hence, the magnetic moment of the loop can be defined as

m = IA

Where m = magnetic moment of the current loop

I = current flowing through the loop

A = direction of area vector

Class 12 Physics Chapter 4 Notes: Exercises &  Solutions

Subject matter experts exclusively prepare Class 12 Physics Chapter 4 Notes at Extramarks keeping in mind the latest CBSE updates regarding the examination pattern. These solutions are useful for both the teachers and the students, and they can be accessed anywhere without much hassle.

To score good  grades in Physics, students need to learn and practise ent-text exercises and additional exercises to check their level of preparation.  Extramarks provides a list of solutions for each and every exercise to encourage and motivate them to step up their learning experience to excel in academics because  solutions include a detailed explanation of topics and point-wise answers. No wonder, millions of students from different boards completely trust by Extramarks  which has a one stop solution to all your problems. 

Class 12 Physics Chapter 4 Notes which describes with illustrations   electric charges, their effects on conductors and magnetism correlate. Click on the respective links below to find solutions to specific exercise questions. 

Students can refer to the respective exercise to access the NCERT solutions Class 12 Physics Chapter 4 Notes. Students can also explore a repository  of educational content available on the Extramarks’ website. Click on the respective links below to access NCERT Solutions for  all Classes from 1 to 12. 

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Key Features of NCERT Solutions Class 12 Physics Chapter 4 Notes

CBSE Class 12 Physics Chapter 4 Notes  is a useful resource  for quick revision before examinations. Physics is one of the key  subjects for  school assignments, tests,  examinations and competitive exams. such as JEE or NEET, and other technical exams. It plays a crucial role in shaping students’ minds to absorb more complex topics in higher studies,   not just to score good grades in Physics alone.

• The Class 12 Physics Chapter 4 Notes will help students understand each and every  concept  included in the chapter.
• Physics is an important subject required for examinations like JEE, NEET or any other technical examination. Hence while studying Physics, authentic and reliable notes and solutions provided by Extramarks will definitely prove useful.
• Subject matter  experts prepare all Class 12 Physics Chapter 4 Notes at Extramarks   who meticulously follow the latest  CBSE  guidelines and  to make you understand the concepts easily and effectively
• The Class 12 Physics Chapter 4 Notes are written in easy-to-understand language which are reliable and accurate. No wonder students have complete trust and faith in Extramarks.
• A repository  of practice questions, CBSE sample papers, and previous year’s question papers, NCERT solutions Extra questions  to get a better hold on the subject.Extramarks doesn’t believe in rote learning, in fact it follows the latest CBSE guidelines to provide experiential learning to students.
• By referring to Class 12 Physics Chapter 4 Notes, students would be able to solve their queries and doubts if any. They cover various concepts  of the chapter and are considered ideal study material.

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