NCERT Class 12 Physics Chapter 4 Notes, Moving Charges and Magnetism Class 12 Chapter 4 Notes

Have you ever thought how a motorcycle’s engine starts or how a fan spins when activated? The everyday activities we perform are made possible by the movement of electrically charged particles and magnetism- the fundamentals of electric motors and generators.

Our updated and easily accessible NCERT Class 12 Physics Chapter 4 Notes are provided below. Careers360's meticulously crafted Moving Charges and Magnetism. To make difficult subjects easier to understand, these notes simplify them by analyzing key formulas and basic ideas.

Clear and organized notes increase the effectiveness of preparing for board examinations, competitive admission exams like JEE and NEET, and even finishing assignments. Our goal is to improve students' study experiences by giving them access to helpful educational resources that will help them do better on tests.

Also, students can refer,

• NCERT Notes Class 12 Physics
• NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges And Magnetism
• NCERT Exemplar Class 12 Physics Chapter 4 Solutions Moving Charges And Magnetism

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 with velocity v

$\therefore F\alpha qBv\sin \theta \Rightarrow F=kqBv\sin \theta$ [ k is constant]

Where in S.I. system k=1

F=qvBsinθ and

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Magnetic field strength B:

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

$\begin{array}{rl}& q=1,v=1\\ & \sin \theta =1\end{array}$

i.e.,

$\theta ={90}^{\circ }$

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=0^o or 180^o

$\therefore F=0$

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

2. When

$v=0,F=0$

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

3. When $\theta ={90}^{\circ },\sin \theta =1$ then the force will be maximum

$F_{max}=qvB$

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).

$B=\frac{F}{qv\sin \theta }$

When

$q=1C,v=1\mathrm{m}/\mathrm{s},\theta ={90}^{\circ }$
That is $\sin \theta =1,F=1\mathrm{N}$

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

$\begin{array}{rl}& dB\propto I\\ & dB\propto dl\\ & dB\propto \sin \theta \\ & dB\propto \frac{1}{r^2}\end{array}$

After we combine them:

$dB\propto \frac{Idl\sin \theta }{r^2}\Rightarrow dB=k\frac{Idl\sin \theta }{r^2}$

Where

$\begin{array}{rl}& k=\frac{{\mu }_0}{4\pi },{\mu }_0=4\pi \times {10}^{-7}T{A}^{-1}m\\ & \overrightarrow{dB}=\frac{{\mu }_{0}I}{4\pi }\frac{\overrightarrow{d}l\times \overrightarrow{r}}{r^3}\end{array}$

dB will be perpendicular to the plane containing dl

• Biot Savart law applications:

A magnetic field (B) is maintained at the centre of a circular current-carrying coil with a radius of r.
$B=\frac{{\mu }_{0}I}{2r}$

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

$B=\frac{{\mu }_{0}nI}{2r}$

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:

$B=\frac{{\mu }_{0}I}{4\pi a}\times (\sin {\varphi }_1+\sin {\varphi }_2)$

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.

$\begin{array}{rl}& {\varphi }_1={\varphi }_2=\frac{\pi }{2}\\ & \therefore B=\frac{{\mu }_0}{4\pi }\frac{2I}{a}\end{array}$

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

$B=\frac{{\mu }_{0}I}{4a}$

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

$B=\frac{{\mu }_{0}I\theta }{4\pi a}$

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

$\int \mathrm{B}\cdot \mathrm{dl}={\mu }_{o}I$

Magnetic field caused by a current-carrying solenoid,

$B={\mu }_{0}nI$

• Magnetic field caused by a toroid or endless solenoid,

$B={\mu }_{0}nI$

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

$x^2=\frac{2m{v}^2}{qE}y$

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 $v$ producing an angle $\theta$ with B will be a helix. Because the charged particle will not be supplied a force by the component of velocity $v\cos \theta$, the particle will move forward with a fixed velocity in the direction of B . The other component, $v\sin \theta$, will produce the force $F=qBv\sin \theta$, which will provide the necessary centripetal force to the charged particle moving along a circular route with radius r.

Centripetal force =

$\frac{m(v\sin \theta {)}^2}{r}=Bqv\sin \theta$
$\therefore v\sin \theta =\frac{Bqr}{m}$

Angular velocity of rotation =

$w=\frac{v\sin \theta }{r}=\frac{Bq}{m}$

Frequency of rotation =

$v=\frac{\omega }{2\pi }=\frac{Bq}{2\pi m}$

The time period of revolution =

$T=\frac{1}{v}=\frac{2\pi m}{Bq}$

• 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

$\Rightarrow \frac{m{v}^2}{r}=Bqv\Rightarrow \frac{mv}{Bq}=r$

the radius of the circular path

Time for travelling a semicircular path =

$\frac{\pi r}{v}=\frac{\pi m}{Bq}$

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

$\frac{m{v}_0^2}{r}=Bqv\Rightarrow v_0=\frac{Bq{r}_0}{m}$

The maximum kinetic energy of the particle =

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m\left(\frac{Bq{r}_0}{m}\right)=(K.E\cdot {)}_{max}=\frac{B^2{q}^2{r}_0^2}{2m}$

The time period of the oscillating electric field =

$T=\frac{2\pi m}{Bq}$

Cyclotron frequency =

$v=\frac{1}{T}=\frac{Bq}{2\pi m}$

Cyclotron angular frequency =

${\omega }_0=2\pi v=\frac{Bq}{m}$

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

$F=IlB\sin \theta$

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.

$\theta$ be the angle formed by the current direction and the magnetic field.

If $\theta =0^{\circ }$ or ${180}^{\circ },\sin \theta \Rightarrow 0\Rightarrow F=0$

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

If

$\theta ={90}^{\circ },\sin \theta =1$
F will be maximum

$F_{max}=IlB$

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:

$\tau =nBIA\sin \alpha$

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 $\alpha$

• 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

$I=\frac{k}{nBA}\theta =G\theta$

where Galvanometer constant

$\begin{array}{rl}& G=\frac{k}{nBA}\\ & \therefore I\propto \theta \end{array}$

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

$I_s=\frac{\theta }{I}=\frac{nBA}{k}$

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

$V_s=\frac{\theta }{V}=\frac{\theta }{IR}=\frac{nBA}{kR}$

• 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.

$\theta =\frac{nBA}{k}I$

• 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 =

$I_g=\frac{V}{R_g+R}$

Here V is the potential difference across the voltmeter.

$\therefore R=\frac{V}{I_g}-G$

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

$S=\left(\frac{I_g}{I-I_g}\right)R_g$

where Rg be the galvanometer’s resistance.

The effective resistance of the ammeter will be,

$R=\frac{R_g}{R_g+S}$

Importance of NCERT Class 12 Physics Chapter 4 Notes

• Aligned with CBSE Syllabus: Covers all key topics such as Biot-Savart’s Law, Ampere’s Circuital Law, and Lorentz Force as per the official syllabus.
• Boosts Exam Preparation: It helps in scoring well in CBSE board exams by simplifying complex concepts and formulas.
• Strengthens Conceptual Understanding: Enhances clarity on how magnetic fields are generated by moving charges and their applications in devices like motors and galvanometers.
• Essential for Competitive Exams: It provides a solid base for entrance exams like JEE Main, NEET, and other engineering/medical tests.
• Improves Problem-Solving Skills: Includes clear explanations and formulas that help in tackling numerical problems with confidence.
• Great for Quick Revision: Well-organized content makes it ideal for last-minute study and recapping before exams.

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