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

Moving charges produce magnetic fields and experience a magnetic force when they move through an external magnetic field.
This CBSE Class 12 chapter connects charge motion, electric current, magnetic fields, current loops and measuring instruments.

Moving Charges and Magnetism explains how electric current produces a magnetic field and how this field acts on moving charges. The chapter also covers charged-particle motion, current-carrying conductors, circular loops, solenoids and galvanometers.

Use these CBSE Class 12 Physics Revision Notes Chapter 4 for the 2026–27 session. Start with force and direction rules. Then revise field formulas, charged-particle paths, current loops and instrument conversions.

Key Takeaways

• Lorentz force: F = q(E + v × B) gives the total force on a moving charge.
• Circular motion: r = mv/(|q|B) when velocity is perpendicular to the field.
• Long solenoid: B = μ₀nI gives the nearly uniform magnetic field inside it.
• Current-loop torque: τ = NIAB sin θ rotates a loop in a uniform field.

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Magnetic Force and Lorentz Force in Class 12 Physics Chapter 4 Notes

Electric currents and moving charges produce magnetic fields. A charge experiences a magnetic force only when it moves through a magnetic field.

Sources of Magnetic Fields

A stationary electric charge produces an electric field.

A moving electric charge produces:

• An electric field
• A magnetic field

A current-carrying conductor also produces a magnetic field in the surrounding space.

The magnetic field is represented by B. It is a vector quantity and follows the principle of superposition.

The SI unit of magnetic field is tesla.

1 T = 1 N/(A m)

A smaller unit is gauss.

1 G = 10⁻⁴ T

Dot and Cross Convention

Directions perpendicular to the plane of the paper are shown as:

• Dot (•): Vector directed out of the plane
• Cross (×): Vector directed into the plane

The dot resembles the tip of an approaching arrow. The cross resembles the tail of an arrow moving away.

Lorentz Force

A charge q moving with velocity v in electric field E and magnetic field B experiences the Lorentz force:

F = q(E + v × B)

The electric force is:

FE = qE

The magnetic force is:

FB = q(v × B)

Its magnitude is:

FB = |q|vB sin θ

Here, θ is the angle between v and B.

For a positive charge, the force direction follows v × B. For a negative charge, the direction is opposite.

Properties of Magnetic Force

• It acts only on a moving charge.
• It becomes zero when v = 0.
• It becomes zero when v is parallel or antiparallel to B.
• It is maximum when v is perpendicular to B.
• It acts perpendicular to both v and B.
• It changes the direction of velocity.
• It does no work on the charge.
• It does not change the particle’s kinetic energy.

Force on a Current-Carrying Conductor

Consider a straight conductor of vector length l carrying current I in an external magnetic field B.

The force is:

F = I(l × B)

Its magnitude is:

F = IlB sin θ

Here, θ is the angle between the current direction and magnetic field.

Important cases:

• If l is parallel to B, F = 0.
• If l is perpendicular to B, F = IlB.

For a conductor of arbitrary shape:

dF = I(dl × B)

The total force is found by adding the force on all current elements.

Motion of Charged Particles in Magnetic Field Revision Notes

The motion of charged particles in magnetic field depends on the angle between the particle’s velocity and the magnetic field.

Motion Parallel to the Magnetic Field

When velocity is parallel or antiparallel to B:

FB = |q|vB sin 0° = 0

The particle moves in a straight line with constant velocity.

Its speed and kinetic energy remain unchanged.

Circular Motion Perpendicular to the Field

When velocity is perpendicular to B, magnetic force acts as centripetal force.

Magnetic force:

FB = |q|vB

Centripetal force:

Fc = mv²/r

Equating both:

mv²/r = |q|vB

Therefore:

r = mv/(|q|B)

The radius increases with mass and speed. It decreases with charge magnitude and magnetic-field strength.

Angular Frequency

For circular motion:

v = ωr

Using r = mv/(|q|B):

ω = |q|B/m

The time period is:

T = 2π/ω

Therefore:

T = 2πm/(|q|B)

The frequency is:

ν = 1/T

ν = |q|B/(2πm)

The frequency and time period are independent of the particle’s speed in non-relativistic motion.

Helical Motion

If velocity has components both parallel and perpendicular to B:

• v⊥ produces circular motion.
• v∥ remains unchanged.
• The particle follows a helical path.

The radius of the helix is:

r = mv⊥/(|q|B)

The time period is:

T = 2πm/(|q|B)

Pitch is the distance travelled along the field in one revolution.

p = v∥T

Therefore:

p = 2πmv∥/(|q|B)

Cyclotron Frequency in Moving Charges and Magnetism Class 12 Notes

The cyclotron frequency is the frequency of circular motion of a charged particle in a uniform magnetic field.

It is given by:

νc = |q|B/(2πm)

The corresponding angular frequency is:

ωc = |q|B/m

The frequency does not depend on:

• Particle speed
• Radius of the circular path
• Kinetic energy in the non-relativistic range

This property allows an alternating electric field to accelerate a charged particle repeatedly.

Basic Principle of a Cyclotron

A cyclotron uses:

• A uniform magnetic field
• Two hollow D-shaped conductors
• An alternating electric field

The magnetic field bends the charged particle into semicircular paths.

The alternating electric field accelerates it whenever it crosses the gap between the two dees.

As the speed increases, the radius increases:

r = mv/(|q|B)

Limitation of Cyclotron Frequency

The formula assumes that the particle’s mass remains constant.

At very high speeds, relativistic effects increase the effective mass. The particle then loses synchronisation with the alternating electric field.

A cyclotron also cannot accelerate neutral particles because magnetic force requires electric charge.

Biot-Savart Law Class 12 Physics Notes

The Biot-Savart law gives the magnetic field produced by a small current element.

Statement and Formula

Consider a current element I dl and a point at displacement r from it.

The magnetic field contribution is:

dB = (μ₀/4π) × I(dl × r)/r³

Using the unit vector r̂:

dB = (μ₀/4π) × I(dl × r̂)/r²

Its magnitude is:

dB = (μ₀/4π) × I dl sin θ/r²

Here:

• I is current.
• dl is the current element.
• r is the distance from the element.
• θ is the angle between dl and r.
• μ₀ is the permeability of free space.

μ₀ = 4π × 10⁻⁷ T m A⁻¹

Dependence of Magnetic Field

According to the law:

dB ∝ I

dB ∝ dl

dB ∝ sin θ

dB ∝ 1/r²

The field contribution is:

• Maximum when θ = 90°
• Zero when θ = 0° or 180°

Direction of Magnetic Field

The direction is given by dl × r̂.

It is perpendicular to the plane containing dl and r.

The right-hand screw rule helps determine the direction.

Comparison with Coulomb’s Law

Magnetic Field Due to Circular Loop Chapter 4 Notes

Every current element in a loop contributes to the magnetic field. On the axis, transverse components cancel and axial components add.

Magnetic Field Due to Circular Loop on Its Axis

For a circular loop of radius R carrying current I, the field at an axial point at distance x is:

B = μ₀IR²/[2(R² + x²)³ᐟ²]

The field is directed along the axis.

Its direction follows the right-hand thumb rule.

Curl the fingers of the right hand in the current direction. The thumb gives the field direction along the axis.

Field at the Centre of the Loop

At the centre:

x = 0

Therefore:

B = μ₀I/(2R)

The field increases with current and decreases with radius.

Coil with N Turns

For a closely wound coil having N turns:

B = μ₀NIR²/[2(R² + x²)³ᐟ²]

At the centre:

B = μ₀NI/(2R)

The field is directly proportional to the number of turns.

Magnetic Field Due to a Circular Arc

For a circular arc of angle φ radians:

B = μ₀Iφ/(4πR)

For a semicircle:

φ = π

Therefore:

B = μ₀I/(4R)

The direction follows the right-hand rule.

Ampere’s Circuital Law and Magnetic Field Due to Current-Carrying Conductor

Ampere’s circuital law relates the magnetic field around a closed path to the current enclosed by that path.

Ampere’s Circuital Law

The law is:

∮B·dl = μ₀Ienclosed

The closed path is called an Amperian loop.

For a symmetric case where B remains constant and tangential over length L:

BL = μ₀Ienclosed

The law is most useful when the current distribution has high symmetry.

Magnetic Field Due to Current-Carrying Conductor

For a long straight conductor carrying current I, choose a circular Amperian loop of radius r.

The magnetic field has the same magnitude at every point on the loop.

B(2πr) = μ₀I

Therefore:

B = μ₀I/(2πr)

Thus:

B ∝ I

B ∝ 1/r

The magnetic-field lines are concentric circles around the conductor.

Right-Hand Thumb Rule

Hold the conductor in the right hand with the thumb pointing in the current direction.

The curled fingers show the magnetic-field direction around the wire.

Field Inside and Outside a Thick Wire

Consider a wire of radius a carrying uniformly distributed current I.

Outside the wire, r ≥ a:

B = μ₀I/(2πr)

Therefore:

B ∝ 1/r

Inside the wire, r < a:

The enclosed current is:

Ienclosed = I(r²/a²)

Using Ampere’s law:

B = μ₀Ir/(2πa²)

Therefore:

B ∝ r

The field increases linearly inside the wire. It decreases as 1/r outside.

Magnetic Field of a Solenoid Class 12 Physics Revision Notes

A solenoid is a closely wound helical conductor. Each turn acts like a circular current loop.

Magnetic Field Inside a Long Solenoid

Let:

• n be the number of turns per unit length.
• I be the current.

The magnetic field of a solenoid is:

B = μ₀nI

If the solenoid has N turns and length l:

n = N/l

Therefore:

B = μ₀NI/l

Nature of the Magnetic Field

Inside a long solenoid:

• The field is strong.
• The field is nearly uniform.
• The field is parallel to the solenoid axis.

Outside a long solenoid:

• The field is very weak.
• It is taken as approximately zero for an ideal solenoid.

Direction of the Field

Curl the fingers of the right hand in the current direction around the turns.

The thumb points towards the north-pole end of the solenoid.

A current-carrying solenoid behaves like a bar magnet.

Force Between Parallel Currents in Moving Charges and Magnetism

Each current-carrying wire creates a magnetic field. This field acts on a nearby current-carrying wire.

Force Between Parallel Currents

Consider two long parallel wires separated by distance d.

Let their currents be I₁ and I₂.

The magnetic field produced by wire 1 at wire 2 is:

B₁ = μ₀I₁/(2πd)

The force on length l of wire 2 is:

F = I₂lB₁

Therefore:

F = μ₀I₁I₂l/(2πd)

The force per unit length is:

F/l = μ₀I₁I₂/(2πd)

The force increases with current and decreases with separation.

Attraction and Repulsion

• Parallel currents in the same direction attract.
• Parallel currents in opposite directions repel.

The forces on both wires are equal in magnitude and opposite in direction.

Torque on a Current Loop Class 12 Physics Notes

A current loop in a uniform magnetic field experiences zero net force. However, it may experience a turning effect.

Torque on a Current Loop

Consider a rectangular loop of area A carrying current I.

If θ is the angle between the area vector and magnetic field:

τ = IAB sin θ

For a coil with N turns:

τ = NIAB sin θ

In vector form:

τ = m × B

The torque on a current loop is:

• Zero when θ = 0° or 180°.
• Maximum when θ = 90°.

Area Vector

The area vector is perpendicular to the plane of the loop.

Its direction is found using the right-hand rule.

Curl the fingers in the current direction. The thumb gives the area-vector direction.

Magnetic Dipole Moment

The magnetic dipole moment of one current loop is:

m = IA

For N turns:

m = NIA

Its SI unit is A m².

The torque becomes:

τ = m × B

Its magnitude is:

τ = mB sin θ

Equilibrium Positions

Stable equilibrium:

m is parallel to B.

θ = 0°

τ = 0

Unstable equilibrium:

m is antiparallel to B.

θ = 180°

τ = 0

The field tends to align the magnetic dipole moment along itself.

Current Loop as a Magnetic Dipole

At large distances, a current loop produces a field similar to a magnetic dipole.

For a loop of area A carrying current I:

m = IA

The magnetic field decreases as 1/r³ at large distances.

Moving Coil Galvanometer Revision Notes

A moving coil galvanometer measures small electric currents. It works on the torque experienced by a current-carrying coil in a magnetic field.

Principle and Construction

The instrument contains:

• A coil with N turns
• Coil area A
• A strong radial magnetic field B
• A soft-iron cylindrical core
• A spring with torsional constant k
• A pointer and scale

The radial field keeps sin θ equal to 1.

Working

The magnetic torque is:

τm = NIAB

The spring provides restoring torque:

τr = kφ

At equilibrium:

kφ = NIAB

Therefore:

φ = NIAB/k

Hence:

φ ∝ I

The angular deflection is directly proportional to current.

Current Sensitivity

Current sensitivity is the deflection per unit current.

SI = φ/I

Therefore:

SI = NAB/k

It increases when:

• N increases.
• A increases.
• B increases.
• k decreases.

Voltage Sensitivity

If the galvanometer resistance is G:

V = IG

Voltage sensitivity is:

SV = φ/V

Therefore:

SV = NAB/(kG)

Increasing the number of turns may increase current sensitivity. It may also increase resistance, so voltage sensitivity may not rise by the same factor.

Conversion into an Ammeter

An ammeter must have very low resistance and is connected in series.

A low shunt resistance S is connected parallel to the galvanometer.

Let:

• G be galvanometer resistance.
• Ig be full-scale galvanometer current.
• I be the required ammeter range.

The potential difference across G and S is equal:

IgG = (I − Ig)S

Therefore:

S = IgG/(I − Ig)

Most current passes through the shunt.

Conversion into a Voltmeter

A voltmeter must have very high resistance and is connected in parallel.

A high resistance R is connected in series with the galvanometer.

For required voltage range V:

V = Ig(G + R)

Therefore:

R = V/Ig − G

This limits the current drawn by the instrument.

Ammeter and Voltmeter Comparison

Moving Charges and Magnetism Formula Notes

Important Terms in Class 12 Physics Chapter 4

Access CBSE Class 12 Physics Revision Notes Chapter 4 in 30 Minutes

Use this 30-minute sequence for quick revision:

• First 10 minutes: Magnetic force, Lorentz force and particle motion
• Next 10 minutes: Biot-Savart law, circular loop, Ampere’s law and solenoid
• Final 10 minutes: Parallel currents, current-loop torque and galvanometer

Check the charge sign before solving force-direction questions. For field questions, identify the current direction and apply the correct right-hand rule.

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