Moving Charges and Magnetism Class 12 Notes | CBSE Physics Chapter 4

Chapter summary

Moving Charges and Magnetism explains how moving charges and electric currents create magnetic fields and how those fields, in turn, exert forces on the charges and currents. It covers the Lorentz force, circular and helical motion in a field, the cyclotron, the Biot-Savart and Ampere laws, fields of wires, loops and solenoids, the force between parallel currents, and the torque on a current loop that drives the moving coil galvanometer. It is one of the highest-yield Class 12 chapters for NEET, supplying both formula-based numericals and conceptual questions on direction and field geometry.

Chapter notes

Key Concepts

1. Magnetic Force on a Moving Charge

Lorentz Force: F = qv × B = qvB sin θ

  • F = force on charge (N)
  • q = charge (C)
  • v = velocity of charge (m/s)
  • B = magnetic field (Tesla, T)
  • θ = angle between v and B

Force is maximum when θ = 90° and zero when θ = 0° or 180° (charge moving parallel to field).

Direction: given by Fleming’s Left-Hand Rule or the right-hand cross product rule.

Important: Magnetic force does no work on the charge (F ⊥ v always). It only changes direction, not speed.

Motion of a Charged Particle in Magnetic Field

Angle (θ)Path
0° or 180°Straight line (no force)
90°Circle (radius r = mv/(qB))
Between 0° and 90°Helix (spiral)

Radius of circular motion: r = mv/(qB)

Time period: T = 2πm/(qB) - independent of velocity!


2. Biot-Savart Law

The magnetic field dB due to a small current element Idl at a point P at distance r is:

dB = (μ₀/4π) × (Idl × r̂)/r²

where μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space)

Magnetic Field Due to Common Configurations

ConfigurationMagnetic Field
Centre of circular loop (radius R, current I)B = μ₀I/(2R)
On axis of circular loop (at distance x)B = μ₀IR²/[2(R² + x²)^(3/2)]
Infinite straight wire (at distance r)B = μ₀I/(2πr)
Inside a solenoid (n turns/length)B = μ₀nI
Inside a toroid (N total turns, radius r)B = μ₀NI/(2πr)

3. Ampere’s Circuital Law

The line integral of B around any closed loop equals μ₀ times the total current enclosed:

∮ B · dl = μ₀I_enclosed

Useful when there is symmetry (infinite wire, solenoid, toroid).


4. Force on a Current-Carrying Conductor

F = Il × B = BIl sin θ

Direction: Fleming’s Left-Hand Rule

Force Between Two Parallel Current-Carrying Conductors

F/l = μ₀I₁I₂/(2πd)

  • Parallel currents (same direction): attract
  • Anti-parallel currents (opposite direction): repel

Definition of Ampere: 1 Ampere is the current which, when flowing through two infinite parallel conductors 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length.


5. Cyclotron

A device that accelerates charged particles to high energies using electric and magnetic fields.

  • Particle moves in semicircles inside two D-shaped electrodes (dees)
  • Magnetic field provides circular motion; Electric field accelerates at each half-turn
  • Cyclotron frequency: ν = qB/(2πm) - independent of radius and speed
  • Maximum energy: E = q²B²R²/(2m), where R is the radius of the dee
  • Cannot accelerate electrons (too light - relativistic effects) or neutral particles

Important Definitions

TermDefinition
Magnetic field (B)Region where a moving charge or current experiences a force; unit: Tesla (T)
Lorentz forceTotal force on a charge in E and B fields: F = qE + qv × B
Biot-Savart lawGives the magnetic field due to a small current element
Ampere’s circuital law∮B·dl = μ₀I_enclosed around a closed loop
CyclotronDevice that accelerates charged particles using crossed E and B fields

Read the rest of the chapter →Hide the rest ↑

Solved Examples

Example 1

A proton moves with velocity 5 × 10⁶ m/s perpendicular to a magnetic field of 0.2 T. Find the radius of the circular path. (mp = 1.67 × 10⁻²⁷ kg)

Answer: r = mv/(qB) = (1.67 × 10⁻²⁷ × 5 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.2) = 8.35 × 10⁻²¹/3.2 × 10⁻²⁰ = 0.26 m

Example 2

Find the magnetic field at the centre of a circular loop of radius 5 cm carrying 2 A current.

Answer: B = μ₀I/(2R) = (4π × 10⁻⁷ × 2)/(2 × 0.05) = 8π × 10⁻⁷/0.1 = 2.51 × 10⁻⁵ T

Example 3

Two parallel wires 10 cm apart carry currents of 5 A and 10 A in the same direction. Find the force per unit length.

Answer: F/l = μ₀I₁I₂/(2πd) = (4π × 10⁻⁷ × 5 × 10)/(2π × 0.1) = (200π × 10⁻⁷)/(0.2π) = 10⁻⁴ N/m (attractive)

Example 4

A solenoid of length 50 cm has 500 turns and carries 3 A. Find B inside.

Answer: n = 500/0.5 = 1000 turns/m. B = μ₀nI = 4π × 10⁻⁷ × 1000 × 3 = 3.77 × 10⁻³ T ≈ 3.8 mT


Important Questions for Board Exams

1-Mark

  1. State Biot-Savart law.
  2. What is the force on a charge moving parallel to a magnetic field?

3-Mark

  1. Derive the expression for magnetic field at the centre of a current-carrying circular loop.
  2. Using Ampere’s law, derive B inside a solenoid.
  3. Explain the working principle of a cyclotron.

5-Mark

  1. State Biot-Savart law. Derive the expression for B on the axis of a circular current loop.
  2. Derive the force per unit length between two parallel current-carrying conductors. Define the Ampere.

Quick Revision Points

  • F = qvB sin θ; magnetic force ⊥ velocity → does no work
  • Circular path: r = mv/(qB); T = 2πm/(qB) - independent of v
  • Biot-Savart: dB = (μ₀/4π)(Idl sin θ)/r²
  • Centre of loop: B = μ₀I/(2R); Straight wire: B = μ₀I/(2πr)
  • Solenoid: B = μ₀nI; Toroid: B = μ₀NI/(2πr)
  • Ampere’s law: ∮B·dl = μ₀I_enc
  • F between wires: F/l = μ₀I₁I₂/(2πd); same direction → attract
  • Cyclotron: ν = qB/(2πm); can’t accelerate electrons or neutrals

Previous: Ch 3 - Current Electricity
Next: Ch 5 - Magnetism and Matter

🃏 Flash Cards: Moving Charges and Magnetism

Class 12 Physics · Chapter 4 – swipe through all 8 cards to understand the whole chapter.

🧲Start here1/8

Magnetic Force on a Moving Charge

A charge feels a magnetic force only when it moves across the field lines.

F = q v B sin θ

θ is the angle between v and B; a charge at rest feels no magnetic force.

  • Maximum F = qvB when v ⊥ B; zero when v ∥ B
  • Force ⊥ to both v and B → does no work, speed unchanged
  • Direction from right-hand rule for q v × B (reverse for negative q)
Full force2/8

The Lorentz Force

Combine the electric and magnetic effects into one expression.

F = q (E + v × B)

For a straight wire of length L: F = B I L sin θ.

  • Electric part qE acts even on a stationary charge
  • Magnetic part q(v × B) acts only on a moving charge
  • A current-carrying wire is just many moving charges → F = BIL sin θ
Core motion3/8

Circular Motion in a Field

Enter ⊥ to B and the force becomes centripetal, bending the path into a circle.

r = m v / (q B) , T = 2π m / (q B)

T and frequency are independent of speed and radius.

  • r = √(2mK) / (qB), so r ∝ √K
  • If v has a part along B → path is a helix
  • Pitch = v cos θ · T (distance advanced per turn)
🔄Application4/8

The Cyclotron

It accelerates ions using the fact that the time period stays constant.

f = q B / (2π m) , K_max = q2 B2 R2 / (2m)

R = dee radius; the alternating voltage runs at the cyclotron frequency f.

  • Constant period lets the field flip in sync each half-circle
  • Energy is gained only in the gap between the dees
  • Cannot accelerate neutral particles or (relativistically) electrons well
🌀Core law5/8

Biot-Savart & Ampere’s Law

Two ways to find the field produced by a current.

dB = (μ0/4π) · I dl sin θ / r2 ; ∮ B · dl = μ0 I_enc

μ0/4π = 10⁻7 T·m/A; Ampere’s law works for symmetric cases.

  • Long straight wire: B = μ0 I / (2π a), so B ∝ 1/a
  • Field circles the wire by the right-hand grip rule
  • Currents outside the chosen loop add zero to ∮ B · dl
🔁Key fields6/8

Loops, Solenoids & Toroids

Standard magnetic-field results you must recall instantly.

Loop centre: B = μ0 N I / (2R) ; Solenoid: B = μ0 n I

Solenoid uses n (turns per metre); toroid uses total N.

  • On loop axis: B = μ0 N I R2 / [2(R2+x2)^(3/2)]
  • Solenoid field is uniform inside, ~zero outside; half (μ0nI/2) at the end
  • Toroid: B = μ0 N I / (2π r), field only inside the ring
🤝Defines ampere7/8

Force Between Parallel Currents

Each wire sits in the other’s field and feels a force.

F / L = μ0 I1 I2 / (2π d)

1 A gives 2 × 10⁻7 N/m for wires 1 m apart in vacuum.

  • Parallel (same-direction) currents attract
  • Antiparallel (opposite-direction) currents repel
  • Force is mutual and equal even when I1 ≠ I2 (Newton’s 3rd law)
🧭Advanced8/8

Torque, Magnetic Moment & Galvanometer

A current loop behaves as a magnetic dipole and twists in a field.

m = N I A , τ = m B sin θ = N I A B sin θ , U = −m·B

θ = angle between the loop’s normal and B.

  • τ max when loop plane ∥ B; τ = 0 (stable) when plane ⊥ B
  • Galvanometer (radial field): φ ∝ I, sensitivity = N A B / k
  • Ammeter = shunt in parallel; voltmeter = high resistance in series
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📝 Practice Moving Charges and Magnetism — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Q1NEET 2020
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is: (take μ₀ = 4π×10⁻⁷ T·m/A)
Correct answer: D. n = N/L = 100/0.5 = 200 turns/m. B = μ₀ n I = (4π×10⁻⁷)(200)(2.5) = 4π×10⁻⁷×500 = 2000π×10⁻⁷ = 6.28×10⁻⁴ T.
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Q2NEET 2019
Ionised hydrogen atoms and α-particles with the same momenta enter perpendicular to a constant magnetic field B. The ratio of the radii of their paths r_H : r_α will be:
Correct answer: D. r = (p)/(qB). With equal momenta p and equal B, r ∝ (1)/(q). For ionised hydrogen q_H = e, for the α-particle q_α = 2e. So (r_H)/(r_α) = (q_α)/(q_H) = (2e)/(e) = 2 : 1.
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Q3NEET 2019
Two toroids 1 and 2 have total number of turns 200 and 100 respectively with average radii 40 cm and 20 cm. If they carry the same current i, the ratio of the magnetic fields along the two loops B₁ : B₂ is:
Correct answer: A. Inside a toroid B = (μ₀ N i)/(2π r), so (B₁)/(B₂) = (N₁)/(N₂)·(r₂)/(r₁) = (200)/(100)×(20)/(40) = 2×(1)/(2) = 1. Hence B₁ : B₂ = 1 : 1.
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Q4NEET 2018
A metallic rod of mass per unit length 0.5 kg/m is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T acts vertically. The current flowing in the rod to keep it stationary is: (take g = 9.8 m/s²)
Correct answer: D. Balancing along the incline: mgsinθ = BILcosθ, giving I = ((m/L)gtanθ)/(B) = (0.5×9.8×tan30°)/(0.25) = (0.5×9.8×0.577)/(0.25) ≈ 11.32 A.
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Q5NEET 2018
The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is:
Correct answer: A. Voltage sensitivity = current sensitivity / resistance, so R = (current sensitivity)/(voltage sensitivity) = (5 div/mA)/(20 div/V) = (5×10³ div/A)/(20 div/V) = 250 Ω.
🔎 See the full step-by-step solution in the app →
Q6NEET 2016
A long straight wire of radius a carries a steady current I, uniformly distributed over its cross-section. The ratio of the magnetic fields B and B’ at radial distances (a)/(2) and 2a respectively from the axis of the wire is:
Correct answer: B. Inside (r<a): B = (μ₀ I r)/(2π a²), so at r=(a)/(2), B = (μ₀ I)/(4π a). Outside (r>a): B’ = (μ₀ I)/(2π r), so at r=2a, B’ = (μ₀ I)/(4π a). The ratio B : B’ = 1.
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Q7NEET 2016
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57×10⁻² T. If e/m for the electron is 1.76×10¹¹ C/kg, the frequency of revolution of the electron is:
Correct answer: A. Cyclotron frequency f = (qB)/(2π m) = ((e/m)B)/(2π) = (1.76×10¹¹×3.57×10⁻²)/(2π) = (6.28×10⁹)/(6.28) = 1×10⁹ Hz = 1 GHz.
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Q8NEET 2014
Two identical long conducting wires are placed at right angles to each other, just touching at their common point O (one above the other). They carry currents I₁ and I₂. The magnitude of the magnetic field at a point P at perpendicular distance d from O, along a line perpendicular to the plane containing the wires, is:
Correct answer: D. Each wire gives B = (μ₀ I)/(2π d) at P. Because the two wires are perpendicular, their fields at P are mutually perpendicular, so they add in quadrature: B = (μ₀)/(2π d)√(I₁² + I₂²).
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Q9NEET 2011
A current-carrying closed loop in the form of a right-angled isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is:
Correct answer: A. The net force on a closed current loop in a uniform field is zero. The arm AB lies along B, so the force on it is zero (F_(AB)=0). Therefore F_(BC) + F_(AC) = 0, giving F_(AC) = -F_(BC) = -F.
🔎 See the full step-by-step solution in the app →
Q10NEET 1998
Two long parallel wires are at a distance of 1 m. Both carry 1 A of current. The force of attraction per unit length between the two wires is:
Correct answer: A. (F)/(L) = (μ₀ I₁ I₂)/(2π d) = ((2×10⁻⁷)(1)(1))/(1) = 2×10⁻⁷ N/m. Same-direction currents attract.
🔎 See the full step-by-step solution in the app →
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Frequently Asked Questions

What is the Lorentz force?

The Lorentz force is the total force on a charge q moving with velocity v in the presence of both an electric field E and a magnetic field B, given by F = q(E + v x B). The electric part qE acts even on a charge at rest, while the magnetic part q(v x B) acts only on a moving charge and is always perpendicular to its velocity.

What is the formula for the radius and time period of a charged particle moving in a magnetic field?

When a charge enters perpendicular to a uniform field B, the magnetic force becomes centripetal and the path is a circle of radius r = mv / (qB), with time period T = 2 pi m / (qB). The time period and frequency depend only on the charge-to-mass ratio and B, not on the speed or the radius.

How much is Moving Charges and Magnetism weighted in NEET?

It is part of the Class 12 Magnetic Effects of Current unit and is firmly in the NEET syllabus. Magnetism as a whole usually contributes a few questions every year, and this chapter is high-yield because it mixes direct formula numericals (force, radius, fields, torque) with concept questions, so it is worth mastering fully.

What is the difference between the Biot-Savart law and Ampere’s circuital law?

The Biot-Savart law, dB = (mu0 / 4 pi) I dl sin(theta) / r squared, gives the field of a small current element and works for any geometry but needs integration. Ampere’s circuital law, the closed line integral of B dot dl equals mu0 times the enclosed current, is faster but only practical when the situation has high symmetry, such as a long straight wire, solenoid, or toroid.

Why do parallel currents attract while antiparallel currents repel?

Each wire sits in the magnetic field produced by the other and feels a force F per unit length equal to mu0 I1 I2 / (2 pi d). When the currents flow in the same direction the forces pull the wires together, so they attract; when the currents are antiparallel the forces push them apart, so they repel. This mutual force also defines the ampere.

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