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

NCERT Class 12 Physics Chapter 4 Notes: Download PDF

NCERT Class 12 Physics Chapter 4 Notes: Moving Charges and Magnetism offer a simplified and structured way of understanding how electric currents produce magnetic fields and interact with charges. These notes are provided with key concepts, formulas and diagrams so you will be able to learn quickly. All the information is provided in PDF format that can be downloaded for free, perfect for exam revision and practice.

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NCERT Notes for Class 12 Chapter 4 Moving Charges and Magnetism

NCERT Class 12 Physics Chapter 4 Notes depict the relationship between electric currents and the associated magnetic fields in a straightforward way. The important laws, relevant formulas, and applications will guide students through effective revision and exam preparation. Students can use these notes to quickly review for the CBSE board exams, JEE, and NEET exams.

Magnetic Field, Lorentz Force:

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

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

Where in the S.I. system, k=1

F=qvBsinθ and

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

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

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

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 $\theta ={90}^{\circ },\sin \theta =\mathrm{1,}$ 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.

Magnetic Force on a Current-Carrying Conductor

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.

Motion of a Charged Particle in a Uniform Magnetic Field:

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 = Magnetic Lorentz force

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

Radius of the circular path,

$R=\frac{mv}{Bq}$

Angular velocity of rotation,

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

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:

Cyclotron is a technology that we use to accelerate positively charged particles and thereby energise 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.

Biot-Savart’s Law:

The Biot-Savart Law is a fundamental principle in electromagnetism that describes how magnetic fields are generated by electric currents. Because of a current element dl, the strength of magnetic flux density or magnetic field at a point P (dB) will be dependent on

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

After we combine them:

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

Where

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

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.

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

The Magnetic Field on the Axis of the Circular Current-Carrying Loop

In the figure, it is shown that a circular loop of radius R carries a current $I$. Application of Biot-Savart law to a current element of length $dl$ at angular position $\theta$ with the axis of the coil. The current in the segment $d\ell$ causes the field $d\stackrel{¯,}{B}$ which lies in the $x$-y plane as shown below.

Another symmetric $d{\overline{\ell }}^{\prime }$ element that is diametrically opposite to the previously $d\ell$ element causes $d\overrightarrow{B^{\prime }}$.
Due to symmetry, the components of $d\overrightarrow{B}$ and $d\overrightarrow{B^{\prime }}$ perpendicular to the x-axis cancel each other. i.e., these components add to zero.
The x-components of the $d\overrightarrow{B}$ combine to give the total field $\overrightarrow{B}$ at point $P$.

We can use the law of Biot-Savart to find the magnetic field at a point P on the axis of the loop, which is at a distance $x$ from the centre. $d\overline{\ell }$ and $\hat{r}$ are perpendicular and the direction of field $d\overline{B}$ caused by this particular element $d\overline{\ell }$ lies in the x-y plane.

The magnetic field due to the current element is

$\mathrm{dB}=\frac{{\mu }_{\mathrm{o}}\mathrm{I}}{4\pi }\int \frac{\mathrm{d}\mathbf{l}\times \hat{\mathbf{r}}}{{\mathrm{r}}^2}$
Since $r^2=x^2+R^2$
The magnitude $dB$ of the field due to element $d\overline{\ell }$ is:

$dB=\frac{{\mu }_{0}I}{4\pi }\frac{d\ell }{(x^2+R^2)}$
The components of the vector $dB$ are

$\begin{array}{r}d{B}_x=dB\sin \theta =\frac{{\mu }_{0}I}{4\pi }\frac{d\ell }{(x^2+R^2)}\frac{R}{{(x^2+R^2)}^{1/2}}\\ d{B}_y=dB\cos \theta =\frac{{\mu }_{0}I}{4\pi }\frac{d\ell }{(x^2+R^2)}\frac{x}{{(x^2+R^2)}^{1/2}}\end{array}$
Total magnetic field along axis $=B_{\text{axis}}=\int d{B}_x=\int dB\sin \theta$

$\because \int d{B}_y=\int dB\cos \theta =0$
Everything in this expression except $d\overrightarrow{\ell }$ is constant and can be taken outside the integral.

The integral $d\overrightarrow{\ell }$ of is just the circumference of the circle, i.e., $\int d\ell =2\pi R$
So, we get
$\Rightarrow B_{\text{axis}}=\frac{{\mu }_{0}I{R}^2}{2{(x^2+R^2)}^{3/2}}$ (on the axis of a circular loop)
If $x\gg R$, then $B=\frac{{\mu }_{0}I{R}^2}{2x^3}$.
At the centre.

$x=0\Rightarrow B_{\text{centre}}=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi \mathrm{N}i}{R}=\frac{{\mu }_{0}Ni}{2R}=B_{max}$

Ampere Circuital Law:

Ampere's circuital law states, the line integral of the magnetic field $\overrightarrow{B}$ around any closed curve is equal to ${\mu }_0$ times the total current i passing through the area enclosed by the curve.

Mathematical statement: $\oint \overrightarrow{B}d\overrightarrow{l}={\mu }_0\sum i={\mu }_0(i_1+i_3-i_2)$

Also using $\overrightarrow{B}={\mu }_0\overrightarrow{H}\dots ($ where $\overrightarrow{H}=$ magnetising field $)$
$\oint {\mu }_0\overrightarrow{H}\cdot \overrightarrow{dl}={\mu }_0\mathrm{\Sigma }i\Rightarrow \oint \overrightarrow{H}\cdot \overrightarrow{dl}=\mathrm{\Sigma }i$

Fingers are curled in the loop direction, the current in the direction of the thumb is taken as positive, whereas in the direction opposite to that of the thumb is taken as negative.

Now, we can see that the total current crossing the above area is $(i_1+i_3-i_2)$, so any current outside the given area will not be considered. So we have to assume

(Outward $\odot \to +ve$, Inward $\otimes \to -ve$ )

General Guidelines for the Selection of Ampere's path for its Application in Different Situations

(i) Path should be chosen in such a way that at every point of the path, magnetic induction should be either tangential to the path elements or normal to it so that the 'dot' product can be easily handled.

(ii) The path should be chosen in such a way that at every point of the path, magnetic induction is either uniform or zero, so that calculations become easy.

Solenoid

A solenoid is defined as a cylindrical coil of many tightly wound turns of insulated wire with a general diameter of the coil smaller than its length. The solenoid has two ends, and one end behaves like the north pole while the opposite end behaves like the south pole. As the length of the solenoid increases, the interior field becomes more uniform and the external field becomes weaker, which can be seen from the diagram.

As the current flows, a magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of the solenoid.

If the solenoid is of infinite length and the point is well inside the solenoid, the magnetic field is given as
$B_{\text{in}}={\mu }_{0}ni$

Here again, n = number of turns per unit length.

Note - The magnetic field outside the solenoid is zero.

Force between Two Parallel Currents

The properties of force between two parallel current-carrying conductors are as follows-

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.

The magnitude of the force acting per unit length can be given as

$f_{12}=f_{21}=\frac{{\mu }_0{I}_1{I}_2}{2\pi a}$

Torque on a Current Loop

As we have studied the electric dipole in a uniform electric field, it will experience a torque similarly, if we place a rectangular loop carrying a steady current $i$ and placed in a uniform magnetic field experiences a torque. It does not experience a net force.

Let us consider a case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. This is illustrated in the given figure. The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F₁ on it, which is directed into the plane of the loop. Its magnitude is,

$F_1=IbB$

Similarly, it exerts a force F₂ on the arm CD, and F₂ is directed out of the plane of the paper.

$F_2=IbB=F_1$

Thus, the net force on the loop is zero. But these two forces are acting at a distance 'a' between them. This torque on the loop is due to the pair of forces F₁ and F₂. The figure given below shows that the torque on the loop tends to rotate anti-clockwise. This torque is (in magnitude),

$\begin{array}{rl}\tau & =F_1\frac{a}{2}+F_2\frac{a}{2}\\ =IbB\frac{a}{2}+IbB\frac{a}{2}=I(ab)B\\ =IAB\end{array}$

where A = ab is the area of the rectangle.

Now we will discuss the case when the plane of the loop is making an angle $\theta$ with a magnetic field. In the previous case, we have considered $\theta =\frac{\pi }{2}$, but this is now a general case.

Here again, you can see that the forces on arms AB and CD are F₁ and F₂

$F_1=F_2=IbB$

Then the torque will be the

$\begin{array}{rl}\tau & =F_1\frac{a}{2}\sin \theta +F_2\frac{a}{2}\sin \theta \\ =\text{Iab}B\sin \theta \\ =IAB\sin \theta \end{array}$

From the above equations, we can see that the torques can be expressed as the vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as,

$m=IA$

If the coil has N turns, then the magnetic moment formula becomes

$m=NIA$

Its direction is defined by the direction of the Area vector.

So, the Torque equation can be written as,

$\tau =\mathbf{m}\times \mathbf{B}$

The magnitude of the magnetic moment of a current-carrying loop is

$|\overrightarrow{M}|=NiA$

i = current in the loop
N = number of turns in the loop
A = area of cross-section of the loop

Moving Coil Galvanometer:

The Moving Coil Galvanometer 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 the Galvanometer constant

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

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 has 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:

• By connecting a high resistance to a galvanometer, it can be turned into 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,

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

Moving Charges and Magnetism: Previous Year Question and Answer

Q. 1 Biot-Savart law indicates that the moving electrons produce a magnetic field B such that
(1) $B\perp v$
(2) $B\Vert v$
(3) It obeys the inverse cube law
(4) It is along the line joining the electrons and the point of observation

Answer:

According to the Biot-Savart law, the magnitude of $\overrightarrow{B}$ is $B\propto |q|,B\propto v$

$\begin{array}{r}B\propto \sin \varphi ,B\propto \frac{1}{r^2}\\ B\propto \frac{|q|v\sin \varphi }{r^2}\\ B=\frac{{\mu }_0}{4\pi }\frac{|q|v\sin \varphi }{r^2}\end{array}$

Where $\frac{{\mu }_0}{4\pi }$ is the proportionality constant, $r$ ' is the magnitude of the position vector from the charge to the point at which we have to find the magnetic field and $\varphi$ is the angle between $\overrightarrow{v}$ and $\overrightarrow{r}$

B is perpendicular to both v and r.

Hence, the answer is option 1.

Q.2 An electron is projected with uniform velocity along the axis of a current-carrying long solenoid. Which of the following is true?
(1) the electron will be accelerated along the axis
(2) the electron path will be circular about the axis
(3) the electron will experience a force at 45o to the axis and hence execute a helical path
(4) the electron will continue to move with uniform velocity along the axis of the solenoid

Answer:

If the magnetic field of a current-carrying solenoid has a moving electron with uniform velocity across the axis, the electron will face a magnetic force due to the magnetic field determined by $F=-evB\sin {180}^{\circ }=0$ (in case of F=0, the velocity is parallel to the magnetic field or anti-parallel or $\theta =0^{\circ }$ or ${180}^{\circ }$). It will either result in the electron moving with uniform velocity and it will go undeflected across the solenoid axis.

Hence, the answer is option 4.

Q.3 A charged particle q is moving in the presence of a magnetic field B, which is inclined to an angle 30° with the direction of the motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path.

Answer:

The force acting on a charged particle in a magnetic field is given by: $\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

Where
q = charge of the particle
v = velocity of particle and
B = magnetic field in that region

Let velocity have a component along B; this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field.

The motion in a plane perpendicular to B is a circular one, thereby producing a helical motion. Thus, the described path will be helical as shown in the figure.

Importance of Moving charges and magnetism Class 12 Notes

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• NCERT Class 12 Physics Chapter 4 Notes provide a direct overview of Biot-Savart Law, Ampere's Circuital Law, and Lorentz Force, which saves time in case you have to study the day before your exam.

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• The magnetic field due to a current-carrying wire, solenoid, and toroid is covered sufficiently in simple language to make it simple for students to learn difficult ideas.

Collection of Formulas

• Important derivations and formulas are all brought together in one place so you can quickly remember them while solving numericals in the board exam and competitive exams.

Enhancing Problem-Solving Skills

• The notes contain useful examples and solved problems so students can utilise their knowledge in real life, such as the operation of motors, galvanometers, and generators.

Facilitating Preparation for Competitive Exams

• Since this chapter consists of basic physics concepts, the better the notes are prepared, the better the students will retain the conceptual clarity to be successful in JEE, NEET, and other entrance exams.

Visual Learning Aid

• The notes also have some neat diagrams, including magnetic field lines, circular loops, solenoids, and moving charges, which help with visualisation and comprehension of the concepts.

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