NEET Physics · NCERT Class 12

Moving Charges and Magnetism notes

Magnetic Force, Biot-Savart Law, Ampere's Law & Solenoid, Force on Current-Carrying Conductors, Magnetic Dipole & Torque and Moving Coil Galvanometer — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 7 topics
  • 66 min read
  • 32 formulas
  • 62 PYQs
1

Chapter Overview

8 min

Moving Charges and Magnetism explains how moving charges and electric currents produce magnetic effects and experience magnetic forces. The chapter begins with magnetic field, Lorentz force and motion of charged particles in uniform magnetic fields, including circular, helical motion and cyclotron. Biot-Savart law calculates magnetic field due to current elements, straight wires and circular loops. Ampere’s law simplifies field calculations for symmetric systems such as long wires, solenoids and toroids. The chapter then studies force on current-carrying conductors and force between parallel currents. Current loops behave like magnetic dipoles and experience torque in magnetic fields. Finally, the moving coil galvanometer uses torque on a current loop to detect current. For NEET, formulas and direction rules are extremely important.

Key points6

  • Magnetic force is always perpendicular to velocity and magnetic field.
  • A magnetic field does no work on a moving charged particle because force is perpendicular to displacement.
  • Direction rules are crucial: right-hand rule for magnetic force and field, Fleming’s left-hand rule for motor force.
  • Biot-Savart law is useful for finite current distributions; Ampere’s law is powerful for high symmetry.
  • Parallel currents in the same direction attract, while opposite currents repel.
  • Moving coil galvanometer converts current into angular deflection using magnetic torque.

Memory tricks2

Magnetic Force Direction: Use right hand for q(v × B). For negative charge, reverse the direction.

Ampere Law Use: Use Ampere when symmetry is strong: long wire, solenoid, toroid.

Examples2

NEET-Style Snapshot

If a charge enters perpendicular to a uniform magnetic field with doubled speed, circular radius doubles because r = mv/(qB).

Real-Life Example

Electric motors, loudspeakers, galvanometers and cyclotrons all use magnetic force on moving charges or currents.

Common mistakes2

Forgetting Magnetic Force Does No Work

Magnetic force is perpendicular to velocity, so it changes direction of velocity but not speed or kinetic energy.

Using Wrong Direction Rule for Negative Charge

Right-hand rule gives force on positive charge. For electron or negative charge, reverse the force direction.

Reference tables2

Quick revision8

Formulas5

Magnetic Force on Moving Charge
F= qvB sinθF =\text{ qvB sin}θ

Magnitude of force on a charge moving in a magnetic field.

F
Magnetic force
q
Charge
v
Speed of charge
B
Magnetic field
θ
Angle between velocity and magnetic field
Biot-Savart Law
dB=μ0Idl sinθ/(4πr2)dB = μ0\text{Idl sin}θ/(4πr²)

Magnetic field contribution due to a small current element.

dB
Small magnetic field
μ0
Permeability of free space
I
Current
dl
Current element length
r
Distance from current element
Ampere's Circuital Law
∮B⋅dl=μ0Ienclosed∮B·dl = μ0I_{\text{enclosed}}

Line integral of magnetic field around a closed loop equals μ0 times enclosed current.

B
Magnetic field
dl
Small path element
I_enclosed
Current enclosed by Amperian loop
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Diagrams3

2

Magnetic Force

10 min

Magnetic force acts on moving charges in a magnetic field. Its vector form is F = q(v × B), so its magnitude is F = qvB sinθ. The force is perpendicular to both velocity and magnetic field, so it cannot change speed or kinetic energy; it only changes direction. If velocity is perpendicular to magnetic field, the particle moves in a circle with radius r = mv/(qB). If velocity has both parallel and perpendicular components, the perpendicular component causes circular motion while the parallel component remains unchanged, producing helical motion. The cyclotron uses repeated acceleration by electric field and circular motion in magnetic field to accelerate charged particles.

Key points6

  • Magnetic force on a stationary charge is zero.
  • Magnetic force is velocity-dependent unlike electrostatic force.
  • For circular motion, magnetic force acts as centripetal force.
  • Time period in uniform magnetic field is independent of speed for non-relativistic motion.
  • In helical motion, pitch depends on the velocity component parallel to magnetic field.
  • Cyclotron cannot accelerate neutral particles because magnetic force needs charge.

Memory tricks2

Zero Force Case: If velocity is along magnetic field, sin0° = 0, so magnetic force is zero.

Circular Motion: Magnetic force only turns the particle; it does not speed it up.

Examples3

Numerical Example

A proton with speed 2 × 10⁶ m/s enters perpendicular to B = 0.5 T. Radius r = mv/(qB) = 1.67 × 10⁻²⁷ × 2 × 10⁶ /(1.6 × 10⁻¹⁹ × 0.5) ≈ 0.0418 m.

Previous NEET-Type Question

If velocity of a charged particle is parallel to magnetic field, it moves undeflected because magnetic force is zero.

Cyclotron Example

Cyclotron frequency depends on q, B and m, not on radius or speed in the non-relativistic limit.

Common mistakes2

Using qvB When Angle Is Not 90°

Always use F = qvB sinθ unless velocity is clearly perpendicular to magnetic field.

Assuming Magnetic Field Changes Kinetic Energy

Pure magnetic field cannot change kinetic energy because magnetic force does no work.

Reference tables2

Quick revision8

Formulas5

Lorentz Force
F=q(E+v×B)F = q(E + v × B)

Total force on a charge moving in electric and magnetic fields.

F
Total electromagnetic force
q
Charge
E
Electric field
v
Velocity of charge
B
Magnetic field
Magnetic Force Magnitude
F= qvB sinθF =\text{ qvB sin}θ

Force on charge q moving with speed v at angle θ to magnetic field.

F
Magnetic force
q
Magnitude of charge
v
Speed
B
Magnetic field
θ
Angle between v and B
Radius of Circular Motion
r=mv/(qB)r = mv/(qB)

Radius of path when charged particle enters magnetic field perpendicular to velocity.

r
Radius of circular path
m
Mass of particle
v
Speed perpendicular to B
q
Charge
B
Magnetic field
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Diagrams4

3

Biot-Savart Law

9 min

Biot-Savart law gives the magnetic field produced by a small current element. The field contribution is proportional to current, length of current element and sine of the angle between current element and position vector, and inversely proportional to square of distance. Its direction is given by the right-hand screw rule or cross product dl × r. This law is used to derive magnetic field due to a straight current-carrying wire and circular current loop. For a long straight wire, B = μ0I/(2πr). At the centre of a circular loop, B = μ0I/(2R), and for N turns it becomes μ0NI/(2R). NEET often asks field magnitude and direction.

Key points6

  • Biot-Savart law in magnetism resembles Coulomb’s law in electrostatics but includes direction of current element.
  • If θ = 0°, the current element contributes zero field at that point.
  • Magnetic field lines around a straight wire are concentric circles.
  • Increasing current increases magnetic field linearly.
  • For a circular loop, field at centre increases with number of turns.
  • Smaller loop radius gives stronger centre field for the same current.

Memory tricks2

Straight Wire Field: Wire field falls as one over distance: B = μ0I/(2πr).

Loop Field: More turns mean more field: N loops act like N identical contributors.

Examples2

Straight Wire Numerical

For I = 10 A and r = 0.05 m, B = μ0I/(2πr) = 4π × 10⁻⁷ × 10/(2π × 0.05) = 4 × 10⁻⁵ T.

Loop Centre Example

For N = 20, I = 2 A and R = 0.1 m, B = μ0NI/(2R) = 4π × 10⁻⁷ × 20 × 2 / 0.2 ≈ 2.51 × 10⁻⁴ T.

Common mistakes2

Forgetting sinθ in Biot-Savart Law

A current element gives no field along its own line because θ = 0° and sinθ = 0.

Wrong Direction Around Wire

Use right-hand thumb rule: thumb along current, curled fingers show magnetic field direction.

Reference tables2

Quick revision8

Formulas4

Biot-Savart Law
dB=μ0Idl sinθ/(4πr2)dB = μ0\text{Idl sin}θ/(4πr²)

Magnitude of magnetic field due to a small current element.

dB
Small magnetic field
μ0
Permeability of free space
I
Current
dl
Length of current element
θ
Angle between dl and r
r
Distance from current element
Vector Form
dB=(μ0/4π)(Idl×r^)/r2dB = (μ0/4π)(I dl × r̂)/r²

Vector form showing direction using cross product.

dl
Current element vector
r̂
Unit vector from element to observation point
r
Distance to observation point
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Diagrams4

4

Ampere's Law & Solenoid

10 min

Ampere’s circuital law states that the line integral of magnetic field around a closed path equals μ0 times the net current enclosed: ∮B·dl = μ0I. It is especially useful for symmetric current distributions such as an infinite straight wire, long solenoid and toroid. For an infinite straight wire, it gives B = μ0I/(2πr). A solenoid is a long helical coil that produces nearly uniform magnetic field inside and weak field outside; for a long solenoid, B = μ0nI. A toroid is a solenoid bent into a circular ring; its magnetic field is confined mostly inside the core. NEET focuses on field formulas, symmetry and comparison of solenoid and toroid.

Key points6

  • Ampere’s law is always valid, but direct field calculation requires symmetry.
  • For a straight wire, the Amperian loop is a circle centered on the wire.
  • For a solenoid, field inside is strong and approximately uniform.
  • A toroid has no ends, so field leakage is much smaller than a solenoid.
  • Magnetic field direction in solenoid is found by right-hand grip rule.
  • Inside a toroid, field depends on distance from centre as 1/r.

Memory tricks2

Solenoid Formula: Solenoid field depends on turn density, not total length alone: B = μ0nI.

Toroid Trick: Toroid traps the field in a ring; outside field is nearly zero.

Examples2

Solenoid Numerical

A solenoid has n = 1000 turns/m and I = 2 A. Field B = μ0nI = 4π × 10⁻⁷ × 1000 × 2 ≈ 2.51 × 10⁻³ T.

Previous NEET-Type Question

Magnetic field outside an ideal long solenoid is nearly zero, while inside it is uniform and equal to μ0nI.

Common mistakes2

Using Ampere's Law Without Symmetry

Ampere’s law is always true, but simple calculation of B requires a path where B is constant or zero on parts.

Confusing n and N

In solenoid, n is turns per unit length. In toroid formula, N is total number of turns.

Reference tables2

Quick revision8

Formulas4

Ampere's Circuital Law
∮B⋅dl=μ0Ienclosed∮B·dl = μ0I_{\text{enclosed}}

Line integral of magnetic field around a closed path equals μ0 times current enclosed.

B
Magnetic field
dl
Small path element
I_enclosed
Net current passing through loop
Magnetic Field due to Infinite Straight Wire
B=μ0I/(2πr)B = μ0I/(2πr)

Magnetic field at distance r from an infinitely long straight current-carrying wire.

B
Magnetic field
I
Current
r
Distance from wire
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Diagrams4

5

Force on Current-Carrying Conductors

9 min

A current-carrying conductor placed in a magnetic field experiences force because the moving charges inside it experience magnetic force. For a straight conductor of length L carrying current I in magnetic field B, force is F = BIL sinθ. Direction is given by Fleming’s left-hand rule: first finger for magnetic field, middle finger for current and thumb for force. Two parallel current-carrying wires also exert forces on each other because each wire produces a magnetic field that acts on the other. Parallel currents in the same direction attract, while opposite currents repel. This interaction is used to define the ampere in classical electromagnetism and is frequently tested in NEET.

Key points6

  • Current direction means conventional current direction.
  • Magnetic force on a wire is the total force on moving charges inside it.
  • If current or magnetic field is reversed, force direction reverses.
  • Force between parallel wires decreases with separation.
  • The force between two current-carrying wires is equal in magnitude and opposite in direction.
  • This topic is the basis of electric motor action.

Memory tricks2

Parallel Wires: Same currents attract like friends walking together; opposite currents repel.

Fleming Left Hand: FBI order: First finger Field, second finger current, thumb Force.

Examples2

Numerical Example

A 0.5 m wire carries 4 A perpendicular to B = 0.2 T. Force F = BIL = 0.2 × 4 × 0.5 = 0.4 N.

Previous NEET-Type Question

Two long parallel conductors carrying currents in opposite directions repel each other.

Common mistakes2

Using Electron Direction Instead of Conventional Current

In F = BIL sinθ and Fleming’s rule, I is conventional current direction.

Forgetting Angle

Use F = BIL sinθ. If wire is parallel to B, force is zero.

Reference tables2

Quick revision8

Formulas3

Force on Current-Carrying Wire
F= BIL sinθF =\text{ BIL sin}θ

Force on a straight conductor of length L carrying current I in magnetic field B.

F
Magnetic force
B
Magnetic field
I
Current
L
Length of wire in field
θ
Angle between current direction and magnetic field
Vector Form of Wire Force
F=I(L×B)F = I(L × B)

Direction of force from cross product of length vector and magnetic field.

F
Force vector
I
Current
L
Length vector along current
B
Magnetic field vector
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Diagrams4

6

Magnetic Dipole & Torque

10 min

A current loop behaves like a magnetic dipole. Its magnetic dipole moment is M = IA for one turn and M = NIA for N turns, directed perpendicular to the plane of the loop by the right-hand rule. When placed in a uniform magnetic field, the loop experiences no net force but experiences torque τ = MB sinθ = NIAB sinθ, tending to align its magnetic moment with the field. This principle is used in electric motors and moving coil galvanometers. Torque is generally turning effect of force and also follows τ = rF sinθ, where only the perpendicular component of force produces rotation. Potential energy of a magnetic dipole is U = -M·B = -MB cosθ.

Key points6

  • Magnetic dipole moment direction is given by right-hand grip rule.
  • In a uniform magnetic field, a current loop experiences torque but no net force.
  • Torque is maximum when magnetic moment is perpendicular to magnetic field.
  • Torque is zero when magnetic moment is parallel or antiparallel to field.
  • Only perpendicular force component contributes to torque.
  • Electric motor converts electrical energy into mechanical rotation using magnetic torque.

Memory tricks2

Torque Maximum: Torque loves perpendicular: maximum when M is at 90° to B.

Dipole Energy: Stable means aligned: U is minimum when M is parallel to B.

Examples3

Given Torque Numerical

If r = 3 m, F = 6 N and θ = 90°, then τ = rF sinθ = 3 × 6 × sin90° = 18 N m.

Magnetic Torque Example

A coil with N = 50, area 0.02 m², current 2 A in B = 0.1 T at 90° has τ = NIAB = 50 × 2 × 0.02 × 0.1 = 0.2 N m.

Previous NEET-Type Question

A current loop in a uniform magnetic field behaves like a magnetic dipole with moment M = NIA.

Common mistakes2

Confusing Area Vector with Plane of Loop

Magnetic moment is perpendicular to the plane of loop, not along the wire.

Assuming Net Force on Loop in Uniform Field

In uniform magnetic field, a current loop has zero net force but non-zero torque if not aligned.

Reference tables2

Quick revision8

Formulas5

Magnetic Dipole Moment
M=IAM = IA

Magnetic dipole moment of a single current loop.

M
Magnetic dipole moment
I
Current in loop
A
Area of loop
Magnetic Dipole Moment of Coil
M=NIAM = NIA

Magnetic dipole moment of a coil of N turns.

N
Number of turns
I
Current
A
Area of each turn
Torque on Current Loop
τ= MB sinθ= NIAB sinθτ =\text{ MB sin}θ =\text{ NIAB sin}θ

Torque on a current loop in a uniform magnetic field.

τ
Torque
M
Magnetic dipole moment
B
Magnetic field
θ
Angle between M and B
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Diagrams4

7

Moving Coil Galvanometer

10 min

A moving coil galvanometer is an instrument used to detect and measure small currents. It consists of a rectangular coil suspended in a radial magnetic field between concave pole pieces, with a soft iron core to strengthen the field. When current flows through the coil, it experiences magnetic torque τ = NIAB. The suspension fibre provides restoring torque kθ. At equilibrium, NIAB = kθ, so deflection θ is directly proportional to current. This is the principle of the galvanometer. Current sensitivity is deflection per unit current, and voltage sensitivity is deflection per unit voltage. A galvanometer is converted into an ammeter by connecting a low shunt resistance in parallel and into a voltmeter by connecting a high resistance in series.

Key points6

  • Radial magnetic field keeps plane of coil parallel to field, making torque maximum and proportional to current.
  • Soft iron core increases magnetic field and sensitivity.
  • Galvanometer detects small currents, not large currents directly.
  • Ideal ammeter has very low resistance.
  • Ideal voltmeter has very high resistance.
  • Shunt protects galvanometer by bypassing most current.

Memory tricks2

Ammeter Conversion: Ammeter needs Alternate path: add low shunt in parallel.

Voltmeter Conversion: Voltmeter must avoid drawing current: add high resistance in series.

Examples3

Solved Numerical: Ammeter Shunt

A galvanometer has G = 100 Ω and Ig = 1 mA. To convert it into 1 A ammeter, S = IgG/(I - Ig) = 0.001 × 100 / 0.999 ≈ 0.100 Ω.

Solved Numerical: Voltmeter Resistance

For G = 100 Ω, Ig = 1 mA and voltmeter range 10 V, R = V/Ig - G = 10/0.001 - 100 = 9900 Ω.

Quick Revision Note

Increasing N, A or B increases current sensitivity because θ/I = NAB/k.

Common mistakes3

Connecting Ammeter in Parallel

Ammeter has very low resistance and must be connected in series; parallel connection can cause short circuit.

Connecting Voltmeter in Series

Voltmeter has high resistance and must be connected in parallel across the component.

Forgetting Radial Field Importance

Radial field makes torque proportional to current for all deflections, giving a linear scale.

Reference tables2

Quick revision8

Formulas6

Deflecting Torque
τ= NIABτ =\text{ NIAB}

Torque on galvanometer coil in radial magnetic field.

τ
Deflecting torque
N
Number of turns
I
Current through coil
A
Area of coil
B
Magnetic field
Restoring Torque
τ=kθτ = kθ

Restoring torque due to suspension fibre.

k
Torsional constant
θ
Angular deflection
Galvanometer Current
I=kθ/(NAB)I = kθ/(NAB)

Current is proportional to deflection in a moving coil galvanometer.

I
Current through galvanometer
k
Torsional constant
θ
Deflection
N
Number of turns
A
Area of coil
B
Magnetic field
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Diagrams4

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NEET PYQs from Moving Charges and Magnetism

The latest 3 of 62 previous-year questions from this chapter.

  1. NEET 2026Set 11MediumQ1

    The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is:

    • AGraph 1
    • BGraph 2
    • CGraph 3
    • DGraph 4
  2. NEET 2026Set 11MediumQ2

    A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take μ₀ = 4π × 10⁻⁷ T m/A)

    • A2 A, 10 A m²
    • B2.5 A, 20 A m²
    • C2 A, 4 A m²
    • D2.5 A, 2 A m²
  3. NEET 2026Set 11EasyQ3

    A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0–10 A. The shunt required is:

    • A0.01 Ω
    • B0.10 Ω
    • C1.0 Ω
    • D0.001 Ω

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