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Class 12 Chapter 4

Class 12 Physics Chapter 4: Moving Charges and Magnetism

Moving Charges and Magnetism is one of the most important chapters in Class 12 Physics that explains how moving electric charges produce magnetic fields and how magnetic fields influence charged particles and current-carrying conductors. The chapter introduces concepts such as Lorentz Force, Biot–Savart Law, Ampere's Circuital Law, magnetic effects of current and magnetism, magnetic force on current-carrying conductors, motion of charged particles in magnetic fields, cyclotron, and torque on current loops. These concepts form the foundation for Electromagnetic Induction and are frequently tested in JEE Main and JEE Advanced through conceptual and numerical problems.

At ALLEN, Chapter 4: Moving Charges and Magnetism smart study resources are carefully curated by expert faculty to help students grasp fundamental concepts quickly and revise efficiently before exams. The notes combine concise theory, important formulas, solved examples, previous years' question trends, and exam-focused tips to strengthen conceptual clarity and problem-solving skills. Whether you're preparing for JEE Main, JEE Advanced, or school examinations, these notes provide a structured and effective revision resource to help you score with confidence.

1.0Chapter Snapshot

Parameter	Details
Difficulty Level	Moderate to High
JEE Main Weightage	High (Generally 1–2 Questions)
JEE Advanced Weightage	High (2–4 Integrated Concepts)
Key Concepts Covered	Lorentz Force, Biot–Savart Law, Ampere's Circuital Law, Magnetic Force, Cyclotron, Motion of Charged Particles, Torque on Current Loop

2.0Chapter 4 Smart Study Material by ALLEN Experts: Download PDFs

Continue your preparation by exploring these resources after completing Moving Charges and Magnetism. These study materials will help reinforce concepts, improve problem-solving speed, and strengthen exam readiness.

Practice Resources

Resource	Access
Chapter Mock Test Paper
Chapter Mock Test Solutions

Must-Revise Concepts

AC Motor

AC motor questions

Ampere’s Law

Derivation of Biot Savart’s Law

Difference Between Ammeter And Galvanometer

Dipole Unim Magnetic Field

effect of magnet on current carying wire

Effect of Temperature on Magnetic Properties

Electric Potential in electromagnetism

Electromagnetic field

Electromagnetism

Lorentz Transmations

Magnetic Effects Of Current and Magnetism

Magnetic Field Lines

Magnetism Previous Year Questions With Solutions

Magnetometer

Moving Coil Galvanometer

Solenoids

Toroid

Magnetic Effects Of Current and Magnetism

The magnetic effect of current is the phenomenon where an electric current creates a magnetic field around a conductor. This was first observed by Hans Christian Oersted. Laws like Biot-Savart and Ampere’s Circuital Law help us understand how these fields are formed. Magnetic field lines show the direction of the field around the wire. Devices like solenoids and toroids use this effect to generate uniform magnetic fields. Depending on how materials respond to these fields, they are classified as ferromagnetic, paramagnetic, or diamagnetic.

Related Video:

1.0Concept of Field and Oersted Experiment

A region around any physical quantity where another similar physical quantity experiences force or torque is called a field.

2.0Oersted Experiment

When the direction of current in the conductor is reversed, then the deflection of the magnetic needle is also reversed. Increasing the current or moving the needle closer increases its deflection.

3.0Biot Savart Law

The magnetic field at a point is directly proportional to the current, element length, and the a sine of the angle, and inversely proportional to the square of the distance.

$d B = \frac{μ _{0}}{4 π} \frac{I d lS in θ}{r ^{2}}$

$μ_{0} = 4 π \times 1 0^{- 7} T m a^{- 1}$

$μ_{0}$ : Permeability of free space

$d B = \frac{μ _{0}}{4 π} \frac{I ( d l \times r )}{r ^{3}} \Rightarrow Vector Form$

$B = \int d B = \frac{μ _{0}}{4 π} \int \frac{I ( d l \times r )}{r ^{3}} (Integral Form)$

1.0Application of Biot-Savart Law

Magnetic Field due to Thin Wire of Finite Length

Magnetic Field due to Thin Wire of Finite Length

$B = \frac{μ _{0} I}{4 πa} [s in ϕ_{1} + S in ϕ_{2}] inwards$

Magnetic Field due to Infinite Straight Wire

Magnetic Field due to Infinite Straight Wire

$B = \frac{μ _{0} I}{2 πa}$

Magnetic Field due to Semi Infinite Straight Wire

Case-1

$ϕ_{1} \to 9 0^{o} or ϕ_{2} \to 0°$

Magnetic Field due to Semi Infinite Straight Wire

$B = \frac{μ _{0} I}{4 πa}^{-\otimes}$

Case-2.

$ϕ_{1} \to 9 0^{o} or ϕ_{2} \to θ$

$B = \frac{μ _{0} I}{4 πa} (1 + s in θ)^{-\otimes}$

Case-3.

$ϕ_{1} \to 9 0^{o} or ϕ_{2} \to - θ$

Magnetic Effects Of Current and Magnetism

$B = \frac{μ _{0} I}{4 πa} (1 - s in θ)^{-\otimes}$

Case-4.

$ϕ_{1} \to - θ or ϕ_{2} \to 9 0^{0}$

Magnetic Effects Of Current and Magnetism - Applicatons of biot savarts law

$B = \frac{μ _{0} I}{4 πa} (1 - s in θ)^{-\otimes}$

2.0Magnetic Field due to Circular Loop

Direction: Can be obtained by right hand thumb rule -Curl the right-hand fingers along the current; the thumb shows the magnetic field direction.

Direction of the Magnetic Field due to Circular Loop

Magnetic Field due to Circular Coil

For N loop- $B_{O} = \frac{μ _{0} N I}{2 R}$

Magnetic Field due to Circular Arc

$B = \frac{μ _{0}}{4 π} \frac{I α}{R}$

Wire of circuit loop is uniform

$\frac{B _{ab c}}{B _{a d c}} = (\frac{l _{2}}{l _{1}}) (\frac{l _{1}}{l _{2}}) = \frac{1}{1}$

The magnetic field produced by both the arc is at the centre of the circuit loop is equal in intensity and in reverse direction. So $(B_{O})_{n e t} = 0$ (always and it is free from angle of connection of terminals)

Wire of circuit loop is non-uniform

Case-1. Thickness-Same, Material - different

Case-2. Thickness-Different, Material - Same

Wire of circuit loop is non-uniform Thickness-Same Material - different

$(B_{O})_{n e t} = \frac{μ _{O}}{4 r} (I_{1} - I_{2})$

Wire of circuit loop is non-uniform when Thickness-Different Material - Same

$(B_{O})_{n e t} = \frac{μ _{O}}{4 r} (I_{1} - I_{2})$

Note: When current divides in any symmetrical planar loop made with uniform wire, then the magnetic field at the centre due to this loop is ZERO.

Magnetic Field at the Axis of Circular loop

Magnetic Field at the Axis of Circular loop

$B_{P} = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{\frac{3}{2}}} = \frac{B _{o}}{( 1 + \frac{x ^{2}}{R ^{2}} ) ^{\frac{3}{2}}}$

Case-1: At very large distance from centre, $B_{P} = \frac{μ _{O} I R ^{2}}{2 x ^{3}}$

Case-2: Near centre of the ring $B_{P} = \frac{B _{o}}{( 1 + \frac{x ^{2}}{R ^{2}} ) ^{\frac{3}{2}}}$

3.0Ampere’s Circuital Law and it's Utilities

Ampere’s Circuital Law: Line integral of magnetic field along any closed loop is equal to o times the net current crossing the surface bounded by the loop.

$\oint B . d l = μ_{0} I_{e n c}$

Note:

1. It is applicable only for a steady/constant current

2. It can be applied for any distribution of current but, it is applied for symmetric distribution for calculation purposes.

Magnetic Field due to Long Thin wire

Ampere’s Circuital Law and it's Utilities - Magnetic Field due to Long Thin wire

$B = \frac{μ _{0} I}{2 π r}$

Magnetic Field due to Long Thick wire

Ampere’s Circuital Law and it's Utilities - Magnetic Field due to Long Thick wire

$B_{O u t} = \frac{μ _{0} I}{2 π r}$

$B_{s u r} = \frac{μ _{0} I}{2 π R}$

$B_{in} = \frac{μ _{0} I}{2 π R ^{2}} = \frac{μ _{0} J r}{2}$

The magnetic field is maximum at the surface of the wire.

Magnetic Field due to Long Hollow Cylindrical Conductor

Magnetic Field due to Long Hollow Cylindrical Conductor

At point A, $(r < a)$

$I_{e n c} = 0 \Rightarrow B_{in} = 0$

At point B, $(r \geq b)$

$I_{e n c} = I \Rightarrow B_{o u t} = \frac{μ _{0} I}{2 π r}$

At point C, $(b > r \geq a)$

$B_{mi d} = \frac{μ _{0} I}{2 π r} (\frac{r ^{2} - a ^{2}}{b ^{2} - a ^{2}})$

4.0Magnetic field due to Solenoid

It is a wire wound into a helix with insulated turns. The magnetic field inside is strong and aligned along the axis, while outside it is nearly zero. The field direction inside is determined using the right-hand thumb rule.

Magnetic Field Inside a Long Solenoid

$B = μ_{O} n I = μ_{O} (\frac{N}{l}) I$

M.F. outside the solenoid $B = 0$

M.F. at the edges/end points $B = \frac{μ _{O} n I}{2}$

Key Note: A solenoid creates a magnetic field similar to that of a bar magnet, acting as a magnetic dipole and serving as an electromagnet in various devices.

5.0Magnetic field due to Toroid

A toroid can be considered as a ring-shaped closed solenoid also called endless solenoid. Toroid is a solenoid bent in ring shape.

Point A: Magnetic field in the empty space surrounded by toroid $(r < R_{1}), B = 0$

Point B: Magnetic field outside the toroid $(r > R_{2}), B = 0$

Point C: Inside the Toroid $(R_{1} \leq r \leq R_{2}) B = μ_{O} n I$

$n = \frac{N}{2 π r _{m}}$ $[r_{m} = \frac{R _{1} + R _{2}}{2}$ $= Mean Radius of Toroid]$

6.0Motion of Charge in a Magnetic Field

Magnetic Force on Moving Charge

$F = q (v \times B) = q v B s in θ$

Magnetic force depends on angle between $v$ and $B$

Case-1: θ = 0° or 180°

Case-2: θ = 90°

Magnetic Force on Moving Charge at angle 0 or 180 degrees

F=0

Magnetic Force on Moving Charge at an angle of 90 degrees

$F_{ma x} = q v B$

Note: Direction of force can be identified by right hand thumb rule or right hand palm rule.

Motion of a Charged Particle in an Unvarying Magnetic Field

Case 1

θ = 0°/180°

Straight Line Path

Case 2

θ = 90°

Circular Path

Case 3

θ ≠ 0°, 90°, 180°

Helical Path

Case 1:

$θ$ =0°

$F = q v B s in 0° = 0$

Straight Line Path

$θ = 180°$

$F = q v B s in 1850° = 0$

Straight Line Path

Motion of Charged Particle in an Unvarying Magnetic Field

Case 2: Motion of charge particle in uniform transverse magnetic field

$Radius of Curvature, R = \frac{m v}{qB} (∵ p = m v)$

$R = \frac{p}{qB} = \frac{2 m K}{qB} K : Kinetic Energy$

$K.E, K = q V_{acc}$

$R = \frac{2 m q V _{acc}}{qB} \Rightarrow R = \frac{1}{B} \frac{2 m V _{acc}}{q}$

$Time period (T) : T = \frac{2 πm}{qB}$

$Frequency, f = \frac{1}{T} = \frac{qB}{2 πm}$

$Angular frequency, ω = \frac{qB}{m}$

Case-3.Motion of Charge Particle in Oblique Magnetic Field ( 0°,90°,180°)

Radius of Circular Path, $R = \frac{m v s in θ}{qB}$

Time period of circular motion, $T = \frac{2 πm}{qB}$

Pitch of helix (p): The linear distance travelled by the charge particle in one revolution or in one time period along the external magnetic field direction is called 'pitch of helix'. $T = \frac{2 m v cos θ}{qB}$

Lorentz Force: When a charge moves in an electric field (E) anda magnetic field (B), both forces act on it. The combined effect is called the Lorentz force.

$F_{n e t} = q (v \times B) + (q E)$

7.0Magnetic Force on a Current Carrying Wire

$d F = I (d l \times B)$

$F = I (L_{e ff} \times B) L_{e ff}$ is the displacement vector from starting point of current to end

point of current.

8.0Magnetic Interaction Between Two Parallel Current Carrying Wires

Magnetic force per unit length of each conductor is, $\frac{d F _{12}}{d l} = \frac{d F _{21}}{d l} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$

9.0Definition of Ampere

Magnetic force/unit length for both infinite length conductor gives as

$f = \frac{μ _{0} I _{1} I _{2}}{2 π d} = \frac{( 4 π \times 1 0 ^{- 7} ) (( 1 ) 1 )}{2 π ( 1 )} = 2 \times 1 0^{- 7} N / m$

'Ampere is the current which, when passed through each of two parallel infinite long straight conductors placed in free space at a distance of 1 m from each other, produces between them a force of 2 × 10^–7 N/m

10.0Bar Magnet and Magnetic Dipole

Bar Magnet: It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. Two poles are present in a Bar magnet, North Pole and South Pole.

Magnetic Dipole: A magnetic dipole consists of a duo of magnetic poles of equal and opposite strength separated by a small distance.

Example: Magnetic needle, Bar Magnet, Current carrying coil/solenoid etc.

$M = m l$

Vector quantity
Direction: South → North
Unit: $A - m^{2}$ or J/T

Magnetic Moment of Current Carrying Coil (Loop)

Magnetic Moment $M = N I A$

11.0Torque and Force on Magnetic Dipole

Bar Magnet

$τ = M \times B = MB s in θ$

$τ = MB s in θ$

Case-1. If $θ = 90° \Rightarrow τ = MB (ma x im u m)$

Case-2.If $θ = 0° or 180° \Rightarrow τ = 0 (minim u m)$

Coil or Loop

$τ = N I A B s in θ$

Case-1. If $θ = 90° \Rightarrow τ = N I A B (ma x im u m)$

Case-2.If $θ = 0° or 180° \Rightarrow τ = 0 (minim u m)$

12.0Moving Coil Galvanometer

The galvanometer has a coil with many turns, free to rotate in a uniform radial magnetic field. A soft iron core strengthens and radializes the field, while a spiral spring resists the coil's rotation.

A current-bearinga coil in a magnetic field experiences a torque. $τ = N I A B s in θ$
The spring S provides a counter torque C that balances the magnetic torque, $τ^{'} = Cϕ$
In equilibrium, $Cϕ = N I A B \Rightarrow I = \frac{C}{N A B} ϕ$ , $I \propto ϕ$

It means the deflection produced is proportional to the current flowing through the galvanometer.

Current Sensitivity

The deflection produced in the galvanometer when a unit current flows through

Voltage Sensitivity

The deflection per unit voltage applied across the voltmeter.

$C . S . = \frac{ϕ}{I} = \frac{N A B}{C} r a d / A$

$V . S . = \frac{ϕ}{V} = \frac{ϕ}{I R} = \frac{N A B}{CR} (r a d / V)$

13.0Potential Energy of Magnetic Dipole

Work Done in Rotating the Coil in a Uniform Magnetic Field

Work Done in Rotating the Coil in Uniform Magnetic Field

$W = MB (cos θ_{1} - cos θ_{2})$

Potential Energy of the Coil in Uniform Magnetic Field

$U = - MB cos θ$

$U = - M . B$

Case-1

Case-2

Case-3

When $M an d B$ are parallel

$θ$ = 0°

$τ$ = 0

$U_{min} = - MB$

Potential Energy of the Coil in Uniform Magnetic Field

When $M an d B$ are perpendicular

$θ = 90°$

$τ = MB$

U = 0

When $M an d B$ are anti- parallel

$θ = 180°$

$τ = 0$

$U_{ma x} = MB$

14.0Atomic Magnetism

Effective current	Induced Magnetic Field at center	Magnetic Moment
$I = \frac{q v}{2 π r}$	$B_{in} = \frac{μ _{o} q v}{4 π r ^{2}}$	$M = \frac{q v r}{2}$

15.0Angular SHM of Magnetic Dipole

$T = 2 π \frac{I}{MB}$ I:Moment of Inertia = $\frac{M l ^{2}}{12}$

16.0Terminology Used In Magnetism

Magnetizing field or Magnetic Intensity $(H)$ : Field in which a material is placed for magnetization.

$H = \frac{B _{0}}{μ _{O}} = A / m$

Intensity of magnetization ( $I$ ): The intensity of magnetisation is the induced dipole moment per unit volume when a magnetic material is placed in a magnetising field.

$I = \frac{M}{V} \Rightarrow A / m$

Magnetic susceptibility ( $χ_{m}$ ): It is a scalar with no units & dimensions. Physically it represents the ease with which a magnetic material can be magnetised.

$χ_{m} = \frac{I}{H}$

Magnetic Permeability $(μ)$

$μ = \frac{B}{H} = \frac{T o t a l ma g n e t i c f i e l d in s i d e a ma t er ia l}{M a g n e t i z in g F i e l d} = H / m$

Relative Permeability $(μ_{r})$

$μ_{r} = \frac{μ}{μ _{o}}$

It has no units and dimensions.

Relation between $μ$ and $χ_{m}$

$μ_{r} = 1 + χ_{m}$

17.0Magnetic Materials

Property	Diamagnetic	Paramagnetic	Ferromagnetic
Magnetic Susceptibility ( $χ_{m}$ )	Small and negative	Small and positive	Large and positive
Effect of Magnetic Field	Weakly repelled	Weakly attracted	Strongly attracted
Magnetization (M)	Opposes external field	Aligns weakly with external field	Strong alignment; persists even after removing field
Unpaired Electrons	None	Present	Many unpaired electrons
Relative Permeability (μᵣ)	Slightly < 1	Slightly > 1	≫ 1
Effect of Removing Field	Magnetization disappears immediately	Magnetization disappears quickly	Magnetization can persist (due to hysteresis)
Temperature Effect	Negligible	Susceptibility decreases with increase in temperature	Loses magnetism above Curie temperature

18.0Sample Questions on Magnetic Effects of Current and Magnetism

Q-1. Calculate magnetic field at P

Solution:

Here

$I d l ∣∣ r$ So, $S in θ = 0$

Hence, the magnetic field at P is zero.

Q-2. If point ‘P’ lies out-side the line of wire then magnetic field at point ‘P’ will be :

Sample questions on Magnetic Effects Of Current and Magnetism

Solution:

$ϕ_{1} = (9 0^{o} - α_{1}) ϕ_{2} = (9 0^{o} - α_{2})$

$B_{P} = \frac{μ _{0} I}{4 π d} [s in (9 0^{o} - α_{1}) - s in (9 0^{o} - α_{2})] = \frac{μ _{0} I}{4 π d} (cos α_{1} - cos α_{2})$

Q-3. Circular loop is made by a wire of length 7.5m. If current 5A is flowing in the loop, then find the magnetic field at the centre.

Solution:

$2 π R = 7.5 \Rightarrow R = \frac{7.5}{2 π} \Rightarrow B_{c} = \frac{μ _{0} I}{2 R}$ $\Rightarrow \frac{4 π \times 1 0 ^{- 7} \times 5}{2 \times \frac{7.5}{2 π}} = \frac{40 π ^{2} \times 1 0 ^{- 7}}{15} T$

Q-4. Two symmetrical current carrying rings are placed perpendicular to each other with a common centre. If magnetic field at the centre due to one coil is B, then find a net magnetic field.

Example problems on Magnetic Effects Of Current and Magnetism

Solution:

$B_{H} = \frac{μ _{0} I}{2 R} = B \hat{j}$

$B_{V} = \frac{μ _{0} I}{2 R} = B \hat{i}$

$B_{net} = B_{H}^{2} + B_{V}^{2} \Rightarrow B 2$

Q-5.A hollow cylindrical wire carries a current 5A, having inner and outer radii 'R' and 2R respectively. The magnetic field at a point which is 3R/2 distance away from its axis is (R = 5m).

Solution:

practice problems on Magnetic Effects Of Current and Magnetism

Field inside cross-section of conductor

$B_{P} = \frac{μ _{0} I}{2 π r} (\frac{r ^{2} - a ^{2}}{b ^{2} - a ^{2}})$

$[a = R, b = 2 R, r = \frac{3 R}{2}]$

$B_{P} = \frac{μ _{0} I}{2 π ( \frac{3 R}{2} )} (\frac{( \frac{3 R}{2} ) ^{2} - R ^{2}}{( 2 R ) ^{2} - R ^{2}}) = \frac{5}{36} \cdot \frac{μ _{0} I}{π R} = \frac{5}{36} \cdot \frac{5 \times μ _{0}}{5 \times π} = \frac{5}{9} \times 1 0^{- 7} T$

Also Read:-

Moving Coil Galvanometer	Difference Between Ammeter And Galvanometer	Forces Between Two Parallel Current Carrying Conductors Questions
Curie-Weiss Law	Difference Between Ammeter And Voltmeter	Derivation of Biot savart's Law
Dipole Uniform Magnetic Field	Curie constant	Electric Charges And Fields

1.0Learning Outcomes

After completing these Moving Charges and Magnetism Revision Notes, you will be able to:

Understand the relationship between moving charges and magnetic fields.
Apply Lorentz Force to determine the motion of charged particles in magnetic fields.
Solve numerical problems using the Biot–Savart Law.
Apply Ampere's Circuital Law to calculate magnetic fields around current-carrying conductors.
Calculate the magnetic force acting on current-carrying wires and charged particles.
Analyze the circular and helical motion of charged particles in uniform magnetic fields.
Understand the working principle and applications of the cyclotron.
Calculate torque acting on current loops placed in magnetic fields.
Strengthen problem-solving skills by mastering concepts and formulas frequently tested in JEE Main and JEE Advanced.

2.05-Minute Quick Revision

If you have only 5 minutes before your exam or mock test, focus on revising these high-weightage concepts and formulas to maximize your score.

Must-Revise Concepts

Lorentz Force
Magnetic Force on a Current-Carrying Conductor
Biot–Savart Law
Ampere's Circuital Law
Magnetic Field due to Straight Wire
Magnetic Field due to Circular Loop
Motion of Charged Particle in Magnetic Field
Cyclotron
Torque on Current Loop

Important Formula Recall

Before attempting numerical questions, quickly revise these essential formulas:

Lorentz Force
Magnetic Force on Current-Carrying Conductor
Biot–Savart Law
Ampere's Circuital Law
Magnetic Field due to Straight Wire
Magnetic Field due to Circular Loop
Radius of Circular Motion
Time Period and Frequency of Charged Particle
Cyclotron Frequency
Torque on Current Loop

3.0High Weightage Topics

The following topics have consistently appeared in JEE Main and JEE Advanced examinations and should be given higher priority during revision.

Topic	Focus Area
Lorentz Force	Force on moving charges
Motion in Magnetic Field	Circular and helical motion
Biot–Savart Law	Magnetic field calculations
Ampere's Circuital Law	Infinite wire and symmetry
Magnetic Force on Conductor	Current-carrying wires
Circular Current Loop	Magnetic field at the centre and axis
Cyclotron	Principle and numerical problems
Torque on Current Loop	Magnetic dipole moment

4.0Formula Handbook

The following formulas are the most important for solving JEE Main and JEE Advanced questions from Moving Charges and Magnetism.

(Formula section continues here.)

5.0Common Mistakes & JEE Tips

Moving Charges and Magnetism is a concept-intensive chapter where students commonly lose marks due to incorrect application of vector directions, right-hand rules, and magnetic force equations. Avoiding these mistakes can significantly improve your accuracy in JEE Main and JEE Advanced.

Common Mistake	Correct Approach
Confusing electric field and magnetic field	Magnetic force acts only on moving charges, unlike electric force.
Applying the wrong right-hand rule	Use the appropriate right-hand thumb or Fleming's left-hand rule depending on the problem.
Ignoring the angle between velocity and magnetic field	Magnetic force depends on the sine of the angle between v and B.
Confusing Biot–Savart Law with Ampere's Law	Choose the method based on the symmetry of the current distribution.
Incorrect direction of magnetic force	Determine direction using vector cross product carefully.
Forgetting centripetal force relation	Magnetic force provides the centripetal force during circular motion.
Incorrect cyclotron calculations	Use the correct radius, frequency, and time period formulas.
Incorrect SI unit conversion	Always use Tesla (T), Ampere (A), Coulomb (C), and Newton (N).

ALLEN Faculty Tips

Always draw the magnetic field and current directions before solving numericals.
Practice right-hand thumb rule and Fleming's left-hand rule regularly.
Memorize standard magnetic field formulas for straight wires, circular loops, and solenoids.
Focus on circular motion of charged particles as it is frequently tested.
Revise cyclotron principles and limitations thoroughly.
Solve previous years' questions to improve speed and conceptual clarity.

6.0PYQ Trend Analysis

Moving Charges and Magnetism is one of the highest-weightage chapters in Class 12 Physics for both JEE Main and JEE Advanced. Questions are primarily numerical and concept-based, with a strong emphasis on Lorentz Force, Biot–Savart Law, Ampere's Circuital Law, and motion of charged particles.

Note: The trend below is based on the analysis of previous years' JEE Main and JEE Advanced question papers. The exact number of questions may vary each year.

Concept	Questions Asked (Last 5 Years)	Average Difficulty
Lorentz Force	5–6	Easy–Moderate
Motion of Charged Particle	7–8	Moderate
Biot–Savart Law	5–6	Moderate
Ampere's Circuital Law	4–5	Moderate
Magnetic Field due to Current Loop	3–4	Moderate
Cyclotron	2–3	Moderate
Torque on Current Loop	3–4	Moderate

Most Common Question Types

Force on moving charged particles.
Circular and helical motion in magnetic fields.
Biot–Savart Law numericals.
Ampere's Circuital Law applications.
Magnetic field due to straight wires and circular loops.
Cyclotron principle and calculations.
Torque on current loops.
Assertion–Reason and concept-based MCQs.

7.0Smart Revision Strategy

If you're short on time, revise the chapter in the following order:

Lorentz Force
Motion of Charged Particles
Biot–Savart Law
Magnetic Field due to Straight Wire
Ampere's Circuital Law
Circular Current Loop
Cyclotron
Torque on Current Loop
Magnetic Dipole Moment

Class 12 Physics Chapters List:- JEE & NEET Revision Notes

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

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Class 12 Physics Chapter 4: Moving Charges and Magnetism

Moving Charges and Magnetism is one of the most important chapters in Class 12 Physics that explains how moving electric charges produce magnetic fields and how magnetic fields influence charged particles and current-carrying conductors. The chapter introduces concepts such as Lorentz Force, Biot–Savart Law, Ampere's Circuital Law, magnetic effects of current and magnetism, magnetic force on current-carrying conductors, motion of charged particles in magnetic fields, cyclotron, and torque on current loops. These concepts form the foundation for Electromagnetic Induction and are frequently tested in JEE Main and JEE Advanced through conceptual and numerical problems.

At ALLEN, Chapter 4: Moving Charges and Magnetism smart study resources are carefully curated by expert faculty to help students grasp fundamental concepts quickly and revise efficiently before exams. The notes combine concise theory, important formulas, solved examples, previous years' question trends, and exam-focused tips to strengthen conceptual clarity and problem-solving skills. Whether you're preparing for JEE Main, JEE Advanced, or school examinations, these notes provide a structured and effective revision resource to help you score with confidence.

1.0Chapter Snapshot

Parameter	Details
Difficulty Level	Moderate to High
JEE Main Weightage	High (Generally 1–2 Questions)
JEE Advanced Weightage	High (2–4 Integrated Concepts)
Key Concepts Covered	Lorentz Force, Biot–Savart Law, Ampere's Circuital Law, Magnetic Force, Cyclotron, Motion of Charged Particles, Torque on Current Loop

2.0Chapter 4 Smart Study Material by ALLEN Experts: Download PDFs

Continue your preparation by exploring these resources after completing Moving Charges and Magnetism. These study materials will help reinforce concepts, improve problem-solving speed, and strengthen exam readiness.

Practice Resources

Resource	Access
Chapter Mock Test Paper
Chapter Mock Test Solutions

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AC Motor

AC motor questions

Ampere’s Law

Derivation of Biot Savart’s Law

Difference Between Ammeter And Galvanometer

Dipole Unim Magnetic Field

effect of magnet on current carying wire

Effect of Temperature on Magnetic Properties

Electric Potential in electromagnetism

Electromagnetic field

Electromagnetism

Lorentz Transmations

Magnetic Effects Of Current and Magnetism

Magnetic Field Lines

Magnetism Previous Year Questions With Solutions

Magnetometer

Moving Coil Galvanometer

Solenoids

Toroid

Magnetic Effects Of Current and Magnetism

The magnetic effect of current is the phenomenon where an electric current creates a magnetic field around a conductor. This was first observed by Hans Christian Oersted. Laws like Biot-Savart and Ampere’s Circuital Law help us understand how these fields are formed. Magnetic field lines show the direction of the field around the wire. Devices like solenoids and toroids use this effect to generate uniform magnetic fields. Depending on how materials respond to these fields, they are classified as ferromagnetic, paramagnetic, or diamagnetic.

Related Video:

1.0Concept of Field and Oersted Experiment

A region around any physical quantity where another similar physical quantity experiences force or torque is called a field.

2.0Oersted Experiment

When the direction of current in the conductor is reversed, then the deflection of the magnetic needle is also reversed. Increasing the current or moving the needle closer increases its deflection.

3.0Biot Savart Law

The magnetic field at a point is directly proportional to the current, element length, and the a sine of the angle, and inversely proportional to the square of the distance.

$d B = \frac{μ _{0}}{4 π} \frac{I d lS in θ}{r ^{2}}$

$μ_{0} = 4 π \times 1 0^{- 7} T m a^{- 1}$

$μ_{0}$ : Permeability of free space

$d B = \frac{μ _{0}}{4 π} \frac{I ( d l \times r )}{r ^{3}} \Rightarrow Vector Form$

$B = \int d B = \frac{μ _{0}}{4 π} \int \frac{I ( d l \times r )}{r ^{3}} (Integral Form)$

1.0Application of Biot-Savart Law

Magnetic Field due to Thin Wire of Finite Length

$B = \frac{μ _{0} I}{4 πa} [s in ϕ_{1} + S in ϕ_{2}] inwards$

Magnetic Field due to Infinite Straight Wire

$B = \frac{μ _{0} I}{2 πa}$

Magnetic Field due to Semi Infinite Straight Wire

Case-1

$ϕ_{1} \to 9 0^{o} or ϕ_{2} \to 0°$

$B = \frac{μ _{0} I}{4 πa}^{-\otimes}$

Case-2.

$ϕ_{1} \to 9 0^{o} or ϕ_{2} \to θ$

$B = \frac{μ _{0} I}{4 πa} (1 + s in θ)^{-\otimes}$

Case-3.

$ϕ_{1} \to 9 0^{o} or ϕ_{2} \to - θ$

$B = \frac{μ _{0} I}{4 πa} (1 - s in θ)^{-\otimes}$

Case-4.

$ϕ_{1} \to - θ or ϕ_{2} \to 9 0^{0}$

$B = \frac{μ _{0} I}{4 πa} (1 - s in θ)^{-\otimes}$

2.0Magnetic Field due to Circular Loop

Direction: Can be obtained by right hand thumb rule -Curl the right-hand fingers along the current; the thumb shows the magnetic field direction.

Magnetic Field due to Circular Coil

For N loop- $B_{O} = \frac{μ _{0} N I}{2 R}$

Magnetic Field due to Circular Arc

$B = \frac{μ _{0}}{4 π} \frac{I α}{R}$

Wire of circuit loop is uniform

$\frac{B _{ab c}}{B _{a d c}} = (\frac{l _{2}}{l _{1}}) (\frac{l _{1}}{l _{2}}) = \frac{1}{1}$

The magnetic field produced by both the arc is at the centre of the circuit loop is equal in intensity and in reverse direction. So $(B_{O})_{n e t} = 0$ (always and it is free from angle of connection of terminals)

Wire of circuit loop is non-uniform

Case-1. Thickness-Same, Material - different

Case-2. Thickness-Different, Material - Same

$(B_{O})_{n e t} = \frac{μ _{O}}{4 r} (I_{1} - I_{2})$

Note: When current divides in any symmetrical planar loop made with uniform wire, then the magnetic field at the centre due to this loop is ZERO.

Magnetic Field at the Axis of Circular loop

$B_{P} = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{\frac{3}{2}}} = \frac{B _{o}}{( 1 + \frac{x ^{2}}{R ^{2}} ) ^{\frac{3}{2}}}$

Case-1: At very large distance from centre, $B_{P} = \frac{μ _{O} I R ^{2}}{2 x ^{3}}$

Case-2: Near centre of the ring $B_{P} = \frac{B _{o}}{( 1 + \frac{x ^{2}}{R ^{2}} ) ^{\frac{3}{2}}}$

3.0Ampere’s Circuital Law and it's Utilities

Ampere’s Circuital Law: Line integral of magnetic field along any closed loop is equal to o times the net current crossing the surface bounded by the loop.

$\oint B . d l = μ_{0} I_{e n c}$

Note:

1. It is applicable only for a steady/constant current

2. It can be applied for any distribution of current but, it is applied for symmetric distribution for calculation purposes.

Magnetic Field due to Long Thin wire

$B = \frac{μ _{0} I}{2 π r}$

Magnetic Field due to Long Thick wire

$B_{O u t} = \frac{μ _{0} I}{2 π r}$

$B_{s u r} = \frac{μ _{0} I}{2 π R}$

$B_{in} = \frac{μ _{0} I}{2 π R ^{2}} = \frac{μ _{0} J r}{2}$

The magnetic field is maximum at the surface of the wire.

Magnetic Field due to Long Hollow Cylindrical Conductor

At point A, $(r < a)$

$I_{e n c} = 0 \Rightarrow B_{in} = 0$

At point B, $(r \geq b)$

$I_{e n c} = I \Rightarrow B_{o u t} = \frac{μ _{0} I}{2 π r}$

At point C, $(b > r \geq a)$

$B_{mi d} = \frac{μ _{0} I}{2 π r} (\frac{r ^{2} - a ^{2}}{b ^{2} - a ^{2}})$

4.0Magnetic field due to Solenoid

It is a wire wound into a helix with insulated turns. The magnetic field inside is strong and aligned along the axis, while outside it is nearly zero. The field direction inside is determined using the right-hand thumb rule.

Magnetic Field Inside a Long Solenoid

$B = μ_{O} n I = μ_{O} (\frac{N}{l}) I$

M.F. outside the solenoid $B = 0$

M.F. at the edges/end points $B = \frac{μ _{O} n I}{2}$

Key Note: A solenoid creates a magnetic field similar to that of a bar magnet, acting as a magnetic dipole and serving as an electromagnet in various devices.

5.0Magnetic field due to Toroid

A toroid can be considered as a ring-shaped closed solenoid also called endless solenoid. Toroid is a solenoid bent in ring shape.

Point A: Magnetic field in the empty space surrounded by toroid $(r < R_{1}), B = 0$

Point B: Magnetic field outside the toroid $(r > R_{2}), B = 0$

Point C: Inside the Toroid $(R_{1} \leq r \leq R_{2}) B = μ_{O} n I$

$n = \frac{N}{2 π r _{m}}$ $[r_{m} = \frac{R _{1} + R _{2}}{2}$ $= Mean Radius of Toroid]$

6.0Motion of Charge in a Magnetic Field

Magnetic Force on Moving Charge

$F = q (v \times B) = q v B s in θ$

Magnetic force depends on angle between $v$ and $B$

Case-1: θ = 0° or 180°

Case-2: θ = 90°

F=0

$F_{ma x} = q v B$

Note: Direction of force can be identified by right hand thumb rule or right hand palm rule.

Motion of a Charged Particle in an Unvarying Magnetic Field

Case 1

θ = 0°/180°

Straight Line Path

Case 2

θ = 90°

Circular Path

Case 3

θ ≠ 0°, 90°, 180°

Helical Path

Case 1:

$θ$ =0°

$F = q v B s in 0° = 0$

Straight Line Path

$θ = 180°$

$F = q v B s in 1850° = 0$

Straight Line Path

Case 2: Motion of charge particle in uniform transverse magnetic field

$Radius of Curvature, R = \frac{m v}{qB} (∵ p = m v)$

$R = \frac{p}{qB} = \frac{2 m K}{qB} K : Kinetic Energy$

$K.E, K = q V_{acc}$

$R = \frac{2 m q V _{acc}}{qB} \Rightarrow R = \frac{1}{B} \frac{2 m V _{acc}}{q}$

$Time period (T) : T = \frac{2 πm}{qB}$

$Frequency, f = \frac{1}{T} = \frac{qB}{2 πm}$

$Angular frequency, ω = \frac{qB}{m}$

Case-3.Motion of Charge Particle in Oblique Magnetic Field ( 0°,90°,180°)

Radius of Circular Path, $R = \frac{m v s in θ}{qB}$

Time period of circular motion, $T = \frac{2 πm}{qB}$

Pitch of helix (p): The linear distance travelled by the charge particle in one revolution or in one time period along the external magnetic field direction is called 'pitch of helix'. $T = \frac{2 m v cos θ}{qB}$

Lorentz Force: When a charge moves in an electric field (E) anda magnetic field (B), both forces act on it. The combined effect is called the Lorentz force.

$F_{n e t} = q (v \times B) + (q E)$

7.0Magnetic Force on a Current Carrying Wire

$d F = I (d l \times B)$

$F = I (L_{e ff} \times B) L_{e ff}$ is the displacement vector from starting point of current to end

point of current.

8.0Magnetic Interaction Between Two Parallel Current Carrying Wires

Magnetic force per unit length of each conductor is, $\frac{d F _{12}}{d l} = \frac{d F _{21}}{d l} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$

9.0Definition of Ampere

Magnetic force/unit length for both infinite length conductor gives as

$f = \frac{μ _{0} I _{1} I _{2}}{2 π d} = \frac{( 4 π \times 1 0 ^{- 7} ) (( 1 ) 1 )}{2 π ( 1 )} = 2 \times 1 0^{- 7} N / m$

'Ampere is the current which, when passed through each of two parallel infinite long straight conductors placed in free space at a distance of 1 m from each other, produces between them a force of 2 × 10^–7 N/m

10.0Bar Magnet and Magnetic Dipole

Bar Magnet: It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. Two poles are present in a Bar magnet, North Pole and South Pole.

Magnetic Dipole: A magnetic dipole consists of a duo of magnetic poles of equal and opposite strength separated by a small distance.

Example: Magnetic needle, Bar Magnet, Current carrying coil/solenoid etc.

$M = m l$

Vector quantity
Direction: South → North
Unit: $A - m^{2}$ or J/T

Magnetic Moment of Current Carrying Coil (Loop)

Magnetic Moment $M = N I A$

11.0Torque and Force on Magnetic Dipole

Bar Magnet

$τ = M \times B = MB s in θ$

$τ = MB s in θ$

Case-1. If $θ = 90° \Rightarrow τ = MB (ma x im u m)$

Case-2.If $θ = 0° or 180° \Rightarrow τ = 0 (minim u m)$

Coil or Loop

$τ = N I A B s in θ$

Case-1. If $θ = 90° \Rightarrow τ = N I A B (ma x im u m)$

Case-2.If $θ = 0° or 180° \Rightarrow τ = 0 (minim u m)$

12.0Moving Coil Galvanometer

The galvanometer has a coil with many turns, free to rotate in a uniform radial magnetic field. A soft iron core strengthens and radializes the field, while a spiral spring resists the coil's rotation.

A current-bearinga coil in a magnetic field experiences a torque. $τ = N I A B s in θ$
The spring S provides a counter torque C that balances the magnetic torque, $τ^{'} = Cϕ$
In equilibrium, $Cϕ = N I A B \Rightarrow I = \frac{C}{N A B} ϕ$ , $I \propto ϕ$

It means the deflection produced is proportional to the current flowing through the galvanometer.

Current Sensitivity

The deflection produced in the galvanometer when a unit current flows through

Voltage Sensitivity

The deflection per unit voltage applied across the voltmeter.

$C . S . = \frac{ϕ}{I} = \frac{N A B}{C} r a d / A$

$V . S . = \frac{ϕ}{V} = \frac{ϕ}{I R} = \frac{N A B}{CR} (r a d / V)$

13.0Potential Energy of Magnetic Dipole

Work Done in Rotating the Coil in a Uniform Magnetic Field

$W = MB (cos θ_{1} - cos θ_{2})$

Potential Energy of the Coil in Uniform Magnetic Field

$U = - MB cos θ$

$U = - M . B$

Case-1

Case-2

Case-3

When $M an d B$ are parallel

$θ$ = 0°

$τ$ = 0

$U_{min} = - MB$

When $M an d B$ are perpendicular

$θ = 90°$

$τ = MB$

U = 0

When $M an d B$ are anti- parallel

$θ = 180°$

$τ = 0$

$U_{ma x} = MB$

14.0Atomic Magnetism

Effective current	Induced Magnetic Field at center	Magnetic Moment
$I = \frac{q v}{2 π r}$	$B_{in} = \frac{μ _{o} q v}{4 π r ^{2}}$	$M = \frac{q v r}{2}$

15.0Angular SHM of Magnetic Dipole

$T = 2 π \frac{I}{MB}$ I:Moment of Inertia = $\frac{M l ^{2}}{12}$

16.0Terminology Used In Magnetism

Magnetizing field or Magnetic Intensity $(H)$ : Field in which a material is placed for magnetization.

$H = \frac{B _{0}}{μ _{O}} = A / m$

Intensity of magnetization ( $I$ ): The intensity of magnetisation is the induced dipole moment per unit volume when a magnetic material is placed in a magnetising field.

$I = \frac{M}{V} \Rightarrow A / m$

Magnetic susceptibility ( $χ_{m}$ ): It is a scalar with no units & dimensions. Physically it represents the ease with which a magnetic material can be magnetised.

$χ_{m} = \frac{I}{H}$

Magnetic Permeability $(μ)$

$μ = \frac{B}{H} = \frac{T o t a l ma g n e t i c f i e l d in s i d e a ma t er ia l}{M a g n e t i z in g F i e l d} = H / m$

Relative Permeability $(μ_{r})$

$μ_{r} = \frac{μ}{μ _{o}}$

It has no units and dimensions.

Relation between $μ$ and $χ_{m}$

$μ_{r} = 1 + χ_{m}$

17.0Magnetic Materials

Property	Diamagnetic	Paramagnetic	Ferromagnetic
Magnetic Susceptibility ( $χ_{m}$ )	Small and negative	Small and positive	Large and positive
Effect of Magnetic Field	Weakly repelled	Weakly attracted	Strongly attracted
Magnetization (M)	Opposes external field	Aligns weakly with external field	Strong alignment; persists even after removing field
Unpaired Electrons	None	Present	Many unpaired electrons
Relative Permeability (μᵣ)	Slightly < 1	Slightly > 1	≫ 1
Effect of Removing Field	Magnetization disappears immediately	Magnetization disappears quickly	Magnetization can persist (due to hysteresis)
Temperature Effect	Negligible	Susceptibility decreases with increase in temperature	Loses magnetism above Curie temperature

18.0Sample Questions on Magnetic Effects of Current and Magnetism

Q-1. Calculate magnetic field at P

Solution:

Here

$I d l ∣∣ r$ So, $S in θ = 0$

Hence, the magnetic field at P is zero.

Q-2. If point ‘P’ lies out-side the line of wire then magnetic field at point ‘P’ will be :

Solution:

$ϕ_{1} = (9 0^{o} - α_{1}) ϕ_{2} = (9 0^{o} - α_{2})$

$B_{P} = \frac{μ _{0} I}{4 π d} [s in (9 0^{o} - α_{1}) - s in (9 0^{o} - α_{2})] = \frac{μ _{0} I}{4 π d} (cos α_{1} - cos α_{2})$

Q-3. Circular loop is made by a wire of length 7.5m. If current 5A is flowing in the loop, then find the magnetic field at the centre.

Solution:

$2 π R = 7.5 \Rightarrow R = \frac{7.5}{2 π} \Rightarrow B_{c} = \frac{μ _{0} I}{2 R}$ $\Rightarrow \frac{4 π \times 1 0 ^{- 7} \times 5}{2 \times \frac{7.5}{2 π}} = \frac{40 π ^{2} \times 1 0 ^{- 7}}{15} T$

Q-4. Two symmetrical current carrying rings are placed perpendicular to each other with a common centre. If magnetic field at the centre due to one coil is B, then find a net magnetic field.

Solution:

$B_{H} = \frac{μ _{0} I}{2 R} = B \hat{j}$

$B_{V} = \frac{μ _{0} I}{2 R} = B \hat{i}$

$B_{net} = B_{H}^{2} + B_{V}^{2} \Rightarrow B 2$

Q-5.A hollow cylindrical wire carries a current 5A, having inner and outer radii 'R' and 2R respectively. The magnetic field at a point which is 3R/2 distance away from its axis is (R = 5m).

Solution:

Field inside cross-section of conductor

$B_{P} = \frac{μ _{0} I}{2 π r} (\frac{r ^{2} - a ^{2}}{b ^{2} - a ^{2}})$

$[a = R, b = 2 R, r = \frac{3 R}{2}]$

$B_{P} = \frac{μ _{0} I}{2 π ( \frac{3 R}{2} )} (\frac{( \frac{3 R}{2} ) ^{2} - R ^{2}}{( 2 R ) ^{2} - R ^{2}}) = \frac{5}{36} \cdot \frac{μ _{0} I}{π R} = \frac{5}{36} \cdot \frac{5 \times μ _{0}}{5 \times π} = \frac{5}{9} \times 1 0^{- 7} T$

Also Read:-

Moving Coil Galvanometer	Difference Between Ammeter And Galvanometer	Forces Between Two Parallel Current Carrying Conductors Questions
Curie-Weiss Law	Difference Between Ammeter And Voltmeter	Derivation of Biot savart's Law
Dipole Uniform Magnetic Field	Curie constant	Electric Charges And Fields

1.0Learning Outcomes

After completing these Moving Charges and Magnetism Revision Notes, you will be able to:

Understand the relationship between moving charges and magnetic fields.
Apply Lorentz Force to determine the motion of charged particles in magnetic fields.
Solve numerical problems using the Biot–Savart Law.
Apply Ampere's Circuital Law to calculate magnetic fields around current-carrying conductors.
Calculate the magnetic force acting on current-carrying wires and charged particles.
Analyze the circular and helical motion of charged particles in uniform magnetic fields.
Understand the working principle and applications of the cyclotron.
Calculate torque acting on current loops placed in magnetic fields.
Strengthen problem-solving skills by mastering concepts and formulas frequently tested in JEE Main and JEE Advanced.

2.05-Minute Quick Revision

If you have only 5 minutes before your exam or mock test, focus on revising these high-weightage concepts and formulas to maximize your score.

Must-Revise Concepts

Lorentz Force
Magnetic Force on a Current-Carrying Conductor
Biot–Savart Law
Ampere's Circuital Law
Magnetic Field due to Straight Wire
Magnetic Field due to Circular Loop
Motion of Charged Particle in Magnetic Field
Cyclotron
Torque on Current Loop

Important Formula Recall

Before attempting numerical questions, quickly revise these essential formulas:

Lorentz Force
Magnetic Force on Current-Carrying Conductor
Biot–Savart Law
Ampere's Circuital Law
Magnetic Field due to Straight Wire
Magnetic Field due to Circular Loop
Radius of Circular Motion
Time Period and Frequency of Charged Particle
Cyclotron Frequency
Torque on Current Loop

3.0High Weightage Topics

The following topics have consistently appeared in JEE Main and JEE Advanced examinations and should be given higher priority during revision.

Topic	Focus Area
Lorentz Force	Force on moving charges
Motion in Magnetic Field	Circular and helical motion
Biot–Savart Law	Magnetic field calculations
Ampere's Circuital Law	Infinite wire and symmetry
Magnetic Force on Conductor	Current-carrying wires
Circular Current Loop	Magnetic field at the centre and axis
Cyclotron	Principle and numerical problems
Torque on Current Loop	Magnetic dipole moment

4.0Formula Handbook

The following formulas are the most important for solving JEE Main and JEE Advanced questions from Moving Charges and Magnetism.

(Formula section continues here.)

5.0Common Mistakes & JEE Tips

Moving Charges and Magnetism is a concept-intensive chapter where students commonly lose marks due to incorrect application of vector directions, right-hand rules, and magnetic force equations. Avoiding these mistakes can significantly improve your accuracy in JEE Main and JEE Advanced.

Common Mistake	Correct Approach
Confusing electric field and magnetic field	Magnetic force acts only on moving charges, unlike electric force.
Applying the wrong right-hand rule	Use the appropriate right-hand thumb or Fleming's left-hand rule depending on the problem.
Ignoring the angle between velocity and magnetic field	Magnetic force depends on the sine of the angle between v and B.
Confusing Biot–Savart Law with Ampere's Law	Choose the method based on the symmetry of the current distribution.
Incorrect direction of magnetic force	Determine direction using vector cross product carefully.
Forgetting centripetal force relation	Magnetic force provides the centripetal force during circular motion.
Incorrect cyclotron calculations	Use the correct radius, frequency, and time period formulas.
Incorrect SI unit conversion	Always use Tesla (T), Ampere (A), Coulomb (C), and Newton (N).

ALLEN Faculty Tips

Always draw the magnetic field and current directions before solving numericals.
Practice right-hand thumb rule and Fleming's left-hand rule regularly.
Memorize standard magnetic field formulas for straight wires, circular loops, and solenoids.
Focus on circular motion of charged particles as it is frequently tested.
Revise cyclotron principles and limitations thoroughly.
Solve previous years' questions to improve speed and conceptual clarity.

6.0PYQ Trend Analysis

Moving Charges and Magnetism is one of the highest-weightage chapters in Class 12 Physics for both JEE Main and JEE Advanced. Questions are primarily numerical and concept-based, with a strong emphasis on Lorentz Force, Biot–Savart Law, Ampere's Circuital Law, and motion of charged particles.

Note: The trend below is based on the analysis of previous years' JEE Main and JEE Advanced question papers. The exact number of questions may vary each year.

Concept	Questions Asked (Last 5 Years)	Average Difficulty
Lorentz Force	5–6	Easy–Moderate
Motion of Charged Particle	7–8	Moderate
Biot–Savart Law	5–6	Moderate
Ampere's Circuital Law	4–5	Moderate
Magnetic Field due to Current Loop	3–4	Moderate
Cyclotron	2–3	Moderate
Torque on Current Loop	3–4	Moderate

Most Common Question Types

Force on moving charged particles.
Circular and helical motion in magnetic fields.
Biot–Savart Law numericals.
Ampere's Circuital Law applications.
Magnetic field due to straight wires and circular loops.
Cyclotron principle and calculations.
Torque on current loops.
Assertion–Reason and concept-based MCQs.

7.0Smart Revision Strategy

If you're short on time, revise the chapter in the following order:

Lorentz Force
Motion of Charged Particles
Biot–Savart Law
Magnetic Field due to Straight Wire
Ampere's Circuital Law
Circular Current Loop
Cyclotron
Torque on Current Loop
Magnetic Dipole Moment

Frequently Asked Questions

How does Moving Charges and Magnetism relate to everyday life?

Why is the right-hand thumb rule important in this chapter?

Which concepts require the most numerical practice in Moving Charges and Magnetism?

How can I improve my accuracy in Moving Charges and Magnetism?

What are the common mistakes students make in this magnetism chapter?

Join ALLEN!

Class 12 Physics Chapter 4: Moving Charges and Magnetism

1.0Chapter Snapshot

2.0Chapter 4 Smart Study Material by ALLEN Experts: Download PDFs

Practice Resources

On this page

Magnetic Effects Of Current and Magnetism

1.0Concept of Field and Oersted Experiment

2.0Oersted Experiment

3.0Biot Savart Law

On this page

1.0Learning Outcomes

2.05-Minute Quick Revision

Must-Revise Concepts

Important Formula Recall

3.0High Weightage Topics

4.0Formula Handbook

5.0Common Mistakes & JEE Tips

ALLEN Faculty Tips

6.0PYQ Trend Analysis

Most Common Question Types

7.0Smart Revision Strategy

On this page

Frequently Asked Questions

How does Moving Charges and Magnetism relate to everyday life?

Why is the right-hand thumb rule important in this chapter?

Which concepts require the most numerical practice in Moving Charges and Magnetism?

How can I improve my accuracy in Moving Charges and Magnetism?

What are the common mistakes students make in this magnetism chapter?

Join ALLEN!

Class 12 Physics Chapter 4: Moving Charges and Magnetism

1.0Chapter Snapshot

2.0Chapter 4 Smart Study Material by ALLEN Experts: Download PDFs

Practice Resources

On this page

Magnetic Effects Of Current and Magnetism

1.0Concept of Field and Oersted Experiment

2.0Oersted Experiment

3.0Biot Savart Law

On this page

1.0Learning Outcomes

2.05-Minute Quick Revision

Must-Revise Concepts

Important Formula Recall

3.0High Weightage Topics

4.0Formula Handbook

5.0Common Mistakes & JEE Tips

ALLEN Faculty Tips

6.0PYQ Trend Analysis

Most Common Question Types

7.0Smart Revision Strategy

On this page