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Moving Charges and Magnetism

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Chapter Notes

Overview
This chapter establishes the intimate relationship between electricity and magnetism, showing that moving charges (currents) produce magnetic fields and experience forces in magnetic fields. The one big idea is that electric currents are the sources of magnetic fields, and these fields exert forces on other moving charges and current-carrying conductors, unifying the two phenomena. Students must understand the Biot-Savart law, Ampere's circuital law, the Lorentz force, and the motion of charged particles in magnetic fields to master electromagnetism.
Key Concepts (18)
Magnetic Field B
A vector field that exerts a force on moving charges and current-carrying conductors. Its SI unit is tesla (T).
e.g. The earth's magnetic field of about 3.6 × 10⁻⁵ T causes a compass needle to align north-south.
Lorentz Force
The total force on a charged particle q moving with velocity v in the presence of both electric field E and magnetic field B is F = q[E + (v × B)].
e.g. An electron moving in a cathode ray tube is deflected by both electric and magnetic fields.
Biot-Savart Law
The magnetic field dB due to a current element I dl at a point at distance r is given by dB = (μ₀/4π) (I dl × r̂)/r².
e.g. Used to calculate the magnetic field on the axis of a circular current loop.
Ampere's Circuital Law
The line integral of the magnetic field B around any closed loop equals μ₀ times the total current enclosed by the loop: ∮ B·dl = μ₀ I.
e.g. Used to find the magnetic field due to a long straight wire: B = μ₀I/(2πr).
Permeability of Free Space (μ₀)
A constant of proportionality in the Biot-Savart law and Ampere's law, with exact value μ₀ = 4π × 10⁻⁷ T m/A.
e.g. It relates magnetic field to current in vacuum.
Right-Hand Thumb Rule (for straight wire)
Grasp the wire with the right hand such that the thumb points in the direction of current; the curled fingers give the direction of the magnetic field.
e.g. For a current flowing upward, the magnetic field circles the wire anticlockwise when viewed from above.
Right-Hand Thumb Rule (for circular loop)
Curl the fingers of the right hand in the direction of current in the loop; the thumb points in the direction of the magnetic field at the centre.
e.g. For a clockwise current, the magnetic field at the centre points into the plane.
Cyclotron
A device that accelerates charged particles using a magnetic field to bend their path and an electric field to increase their energy. The frequency of revolution is independent of energy: ω = qB/m.
e.g. Used to accelerate protons for nuclear physics experiments.
Solenoid
A long coil of wire wound in a helix, producing a uniform magnetic field inside parallel to its axis. The field inside is B = μ₀ n I, where n is the number of turns per unit length.
e.g. An electromagnet used in MRI machines.
Torque on a Current Loop
A rectangular current loop of area A carrying current I in a uniform magnetic field B experiences a torque τ = I A B sin θ, where θ is the angle between the normal to the loop and B.
e.g. This torque is the principle behind the working of a galvanometer.
Magnetic Dipole Moment (m)
For a current loop, m = I A n̂, where n̂ is the unit normal to the plane of the loop. The torque on the loop is τ = m × B.
e.g. A circular loop of area 0.1 m² carrying 2 A has a magnetic moment of 0.2 A m².
Current Element I dl
An infinitesimal segment of a current-carrying conductor, treated as a vector source of magnetic field in the Biot-Savart law.
e.g. A 1 cm segment of a wire carrying 10 A is a current element of magnitude 0.1 A m.
Amperian Loop
An imaginary closed loop used in Ampere's circuital law, chosen to exploit symmetry for easy evaluation of the magnetic field.
e.g. A circle concentric with a long straight wire is an Amperian loop.
Helical Motion
The path of a charged particle moving in a uniform magnetic field when its velocity has a component parallel to the field. The particle moves in a helix with radius r = mv⊥/(qB) and pitch p = v∥ T.
e.g. An electron entering a magnetic field at an angle follows a helical path.
Pitch of a Helix
The distance moved along the magnetic field in one complete revolution: p = v∥ T = 2π m v∥/(qB).
e.g. If v∥ = 10⁶ m/s and T = 10⁻⁷ s, pitch = 0.1 m.
Force Between Parallel Currents
Two parallel current-carrying conductors attract if currents are in the same direction and repel if opposite. Force per unit length: f = μ₀ I₁ I₂/(2π d).
e.g. This force is used to define the ampere: 1 A produces 2 × 10⁻⁷ N/m between two wires 1 m apart.
Tesla (T)
The SI unit of magnetic field: 1 T = 1 N/(A·m). A smaller unit is gauss (1 G = 10⁻⁴ T).
e.g. The earth's magnetic field is about 0.36 G or 3.6 × 10⁻⁵ T.
Gauss
A non-SI unit of magnetic field: 1 G = 10⁻⁴ T.
e.g. A refrigerator magnet has a field of about 100 G.
Formulas & Equations (16)
F = q [E + (v × B)]
Reads as: F equals q times the sum of E and the cross product of v and B
Variables: F = force (N), q = charge (C), E = electric field (N/C), v = velocity (m/s), B = magnetic field (T)
Use when: Calculating total electromagnetic force on a moving charged particle
⚠ Forgetting that the magnetic force is perpendicular to both v and B, not parallel to either
F = I l × B
Reads as: F equals I times the cross product of l and B
Variables: F = force (N), I = current (A), l = length vector of conductor (m), B = magnetic field (T)
Use when: Finding force on a straight current-carrying conductor in a uniform magnetic field
⚠ Using the magnitude of current as a vector; l is the vector, not I
dB = (μ₀/4π) (I dl × r̂)/r²
Reads as: d B equals mu-zero over four pi times I d l cross r-hat over r squared
Variables: dB = magnetic field element (T), μ₀ = 4π × 10⁻⁷ T m/A, I = current (A), dl = length element (m), r̂ = unit vector from element to point, r = distance (m)
Use when: Calculating magnetic field due to a small current element
⚠ Forgetting the cross product direction; dB is perpendicular to both dl and r
B = (μ₀ I R²) / [2 (x² + R²)^(3/2)]
Reads as: B equals mu-zero I R squared over two times x squared plus R squared to the power three-halves
Variables: B = magnetic field on axis (T), μ₀ = 4π × 10⁻⁷ T m/A, I = current (A), R = radius of loop (m), x = distance from centre (m)
Use when: Finding magnetic field on the axis of a circular current loop
⚠ Forgetting the factor of 1/2; at centre (x=0), B = μ₀I/(2R)
B = μ₀ I / (2R)
Reads as: B equals mu-zero I over two R
Variables: B = magnetic field at centre (T), μ₀ = 4π × 10⁻⁷ T m/A, I = current (A), R = radius (m)
Use when: Calculating magnetic field at the centre of a circular loop
⚠ Using diameter instead of radius
∮ B·dl = μ₀ I
Reads as: The line integral of B dot d l equals mu-zero times I
Variables: B = magnetic field (T), dl = path element (m), μ₀ = 4π × 10⁻⁷ T m/A, I = enclosed current (A)
Use when: Finding magnetic field in symmetric situations (long wire, solenoid, toroid)
⚠ Not correctly identifying the enclosed current; using the total current instead of the current through the loop
B = μ₀ I / (2π r)
Reads as: B equals mu-zero I over two pi r
Variables: B = magnetic field (T), μ₀ = 4π × 10⁻⁷ T m/A, I = current (A), r = distance from wire (m)
Use when: Finding magnetic field due to an infinitely long straight current-carrying wire
⚠ Forgetting that this applies only for r > radius of wire; inside wire, B ∝ r
B = (μ₀ I r) / (2π a²)
Reads as: B equals mu-zero I r over two pi a squared
Variables: B = magnetic field inside wire (T), μ₀ = 4π × 10⁻⁷ T m/A, I = total current (A), r = distance from centre (m), a = radius of wire (m)
Use when: Finding magnetic field inside a long straight wire of radius a with uniform current distribution
⚠ Using this formula for r > a; outside wire, use B = μ₀I/(2πr)
B = μ₀ n I
Reads as: B equals mu-zero n I
Variables: B = magnetic field inside solenoid (T), μ₀ = 4π × 10⁻⁷ T m/A, n = number of turns per unit length (m⁻¹), I = current (A)
Use when: Finding magnetic field inside a long solenoid (length >> radius)
⚠ Forgetting that this is for the interior; field outside is nearly zero
f = μ₀ I₁ I₂ / (2π d)
Reads as: f equals mu-zero I-one I-two over two pi d
Variables: f = force per unit length (N/m), μ₀ = 4π × 10⁻⁷ T m/A, I₁, I₂ = currents (A), d = separation (m)
Use when: Calculating force per unit length between two parallel current-carrying conductors
⚠ Using distance between centres of wires incorrectly; d is the perpendicular separation
τ = I A B sin θ
Reads as: tau equals I A B sine theta
Variables: τ = torque (N m), I = current (A), A = area of loop (m²), B = magnetic field (T), θ = angle between normal to loop and B
Use when: Finding torque on a current loop in a uniform magnetic field
⚠ Using angle between plane of loop and B instead of angle between normal and B
τ = m × B
Reads as: tau equals m cross B
Variables: τ = torque (N m), m = magnetic dipole moment (A m²), B = magnetic field (T)
Use when: Expressing torque in vector form for a magnetic dipole
⚠ Forgetting that m = I A n̂, where n̂ is the normal direction
r = m v / (q B)
Reads as: r equals m v over q B
Variables: r = radius of circular path (m), m = mass (kg), v = speed (m/s), q = charge (C), B = magnetic field (T)
Use when: Finding radius of circular motion of a charged particle in a perpendicular magnetic field
⚠ Forgetting that v must be perpendicular to B; if not, use perpendicular component
ω = q B / m
Reads as: omega equals q B over m
Variables: ω = angular frequency (rad/s), q = charge (C), B = magnetic field (T), m = mass (kg)
Use when: Finding cyclotron frequency; independent of speed
⚠ Confusing ω with frequency f; f = ω/(2π)
T = 2π m / (q B)
Reads as: T equals two pi m over q B
Variables: T = time period (s), m = mass (kg), q = charge (C), B = magnetic field (T)
Use when: Finding time for one complete revolution in a magnetic field
⚠ Forgetting that T is independent of speed
p = v∥ T = 2π m v∥ / (q B)
Reads as: p equals v-parallel T equals two pi m v-parallel over q B
Variables: p = pitch (m), v∥ = velocity component parallel to B (m/s), T = time period (s), m = mass (kg), q = charge (C), B = magnetic field (T)
Use when: Finding pitch of helical path of a charged particle in a magnetic field
⚠ Using total speed v instead of v∥
How It Works
Oersted's Experiment
  1. Place a magnetic compass needle near a straight wire carrying current.
  2. Observe that the needle deflects, aligning tangentially to an imaginary circle centred on the wire.
  3. Reverse the current direction; the needle deflects in the opposite direction.
  4. Increase current or bring needle closer; deflection increases.
💡 The deflection shows that moving charges (currents) produce a magnetic field; the field lines are concentric circles around the wire.
Motion of a Charged Particle in a Uniform Magnetic Field
  1. If velocity v is perpendicular to B, the magnetic force q(v × B) acts as centripetal force, causing circular motion.
  2. Equate centripetal force mv²/r to qvB to get radius r = mv/(qB).
  3. Angular frequency ω = qB/m is independent of speed.
  4. If v has a component parallel to B, that component remains unchanged, resulting in helical motion.
  5. Pitch p = v∥ T = 2π m v∥/(qB).
💡 Magnetic force does no work; it only changes direction, not speed. The frequency is independent of energy, which is key for cyclotron operation.
Biot-Savart Law Application: Magnetic Field on Axis of Circular Loop
  1. Consider a circular loop of radius R carrying current I.
  2. Take a current element I dl on the loop.
  3. At an axial point P at distance x, dB = (μ₀/4π) I dl / (x² + R²).
  4. Resolve dB into components: dBx = dB cosθ and dB⊥ = dB sinθ.
  5. By symmetry, dB⊥ components cancel; only dBx adds up.
  6. Integrate dBx over the entire loop: B = (μ₀ I R²) / [2 (x² + R²)^(3/2)].
💡 At the centre (x=0), B = μ₀I/(2R). The direction is given by the right-hand thumb rule for loops.
Ampere's Circuital Law: Magnetic Field Due to a Long Straight Wire
  1. Choose a circular Amperian loop of radius r concentric with the wire.
  2. By symmetry, B is tangential and constant in magnitude on the loop.
  3. Line integral ∮ B·dl = B × 2πr.
  4. Current enclosed I_e = I (for r > a) or I_e = I (r²/a²) (for r < a).
  5. Apply Ampere's law: B × 2πr = μ₀ I_e.
  6. Solve for B: for r > a, B = μ₀I/(2πr); for r < a, B = (μ₀ I r)/(2π a²).
💡 Ampere's law is useful only for symmetric current distributions. The field outside the wire varies as 1/r, inside as r.
Force Between Two Parallel Current-Carrying Conductors
  1. Conductor a produces magnetic field B_a = μ₀ I_a/(2π d) at the location of conductor b.
  2. Conductor b carrying current I_b experiences force F = I_b L × B_a.
  3. Magnitude: F = (μ₀ I_a I_b L)/(2π d).
  4. Direction: using right-hand rule, parallel currents attract, antiparallel repel.
  5. Force per unit length: f = μ₀ I_a I_b/(2π d).
💡 This force is used to define the ampere. Like currents attract, unlike currents repel — opposite to the rule for electric charges.
Torque on a Rectangular Current Loop in a Uniform Magnetic Field
  1. Consider a rectangular loop of sides a and b, area A = ab, carrying current I.
  2. Place it in uniform B such that the normal to the loop makes angle θ with B.
  3. Forces on arms AB and CD: F₁ = F₂ = I b B, but they are opposite and not collinear, forming a couple.
  4. Forces on arms BC and DA cancel along the axis.
  5. Torque magnitude: τ = I A B sin θ.
  6. In vector form: τ = m × B, where m = I A n̂.
💡 Net force on the loop is zero, but torque tends to align the loop's normal with the field. This is the principle of the galvanometer.
Magnetic Field Inside a Solenoid
  1. Consider a long solenoid (length >> radius) with n turns per unit length, carrying current I.
  2. Choose a rectangular Amperian loop abcd: side ab inside, cd outside, bc and ad perpendicular.
  3. Field outside is nearly zero; along bc and ad, B is perpendicular, so B·dl = 0.
  4. Only side ab contributes: ∮ B·dl = B h.
  5. Current enclosed = I × (number of turns through loop) = I (n h).
  6. Ampere's law: B h = μ₀ I n h → B = μ₀ n I.
💡 The field inside is uniform and parallel to the axis; outside, it is negligible. Direction given by right-hand thumb rule for loops.
Exam Traps — Don't Make These Mistakes
⚠ Direction of Magnetic Force on a Moving Charge
✗ The magnetic force is always in the direction of v or B.
→ The magnetic force is perpendicular to both v and B (given by v × B). It is never parallel to either.
✓ The magnetic force is perpendicular to the plane containing v and B, following the right-hand rule for cross products.
⚠ Angle in Torque Formula
✗ Torque on a current loop is I A B cos θ, where θ is the angle between the plane of the loop and B.
→ The correct formula is τ = I A B sin θ, where θ is the angle between the normal to the loop and B. If the plane is parallel to B, θ = 90°, sin 90° = 1.
✓ Torque = I A B sin θ, with θ being the angle between the magnetic field and the normal to the plane of the loop.
⚠ Force Between Parallel Currents: Attraction vs Repulsion
✗ Parallel currents repel each other, like charges.
→ Unlike electric charges, parallel currents attract each other. Antiparallel currents repel.
✓ Parallel currents attract each other; antiparallel currents repel each other.
⚠ Magnetic Field Inside a Current-Carrying Wire
✗ The magnetic field inside a wire is B = μ₀I/(2πr) for all r.
→ That formula applies only outside the wire (r > a). Inside the wire (r < a), the field is B = (μ₀ I r)/(2π a²), which is proportional to r.
✓ For r < a: B = (μ₀ I r)/(2π a²); for r > a: B = μ₀I/(2πr).
⚠ Cyclotron Frequency Dependence on Speed
✗ The cyclotron frequency increases with the speed of the particle.
→ The cyclotron frequency ω = qB/m is independent of speed or energy. Only the radius r = mv/(qB) increases with speed.
✓ The cyclotron frequency ω = qB/m is constant for a given charge and mass in a fixed magnetic field, independent of particle speed.
⚠ Biot-Savart Law vs Coulomb's Law: Angle Dependence
✗ Both laws have the same inverse-square dependence and no angle dependence.
→ Coulomb's law has no angle dependence; Biot-Savart law includes sin θ, where θ is the angle between dl and r.
✓ Biot-Savart law has an additional sin θ factor; the field is zero along the direction of dl (θ = 0).
Compare & Contrast
Nature of source and field direction
Biot-Savart Law
Coulomb's Law
Biot-Savart law involves a vector source (I dl) and produces a field perpendicular to the plane containing dl and r; Coulomb's law involves a scalar source (charge) and produces a field along the line joining source and point.
Force on a stationary charge
Electric Field E
Magnetic Field B
Electric field exerts a force on a stationary charge (F = qE); magnetic field exerts no force on a stationary charge (F = 0 if v = 0).
Nature of force between them
Parallel Currents
Antiparallel Currents
Parallel currents attract each other; antiparallel currents repel each other.
Work done by the field
Motion in Electric Field
Motion in Magnetic Field
Electric field can do work on a charged particle, changing its kinetic energy; magnetic field does no work, only changes direction of motion.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total about 12 marks
Most asked: Biot-Savart law and its application to circular loop · Ampere's circuital law and magnetic field due to long wire and solenoid · Lorentz force and motion of charged particles in magnetic field · Force between two parallel current-carrying conductors · Torque on a current loop in a uniform magnetic field
Don't skip: Derivation of magnetic field on the axis of a circular current loop using Biot-Savart law; Derivation of force between two parallel current-carrying conductors and definition of ampere; Derivation of torque on a rectangular current loop in a uniform magnetic field; Numerical problems on cyclotron motion and helical path
One-Line Revision (read aloud before the exam)
  1. Oersted discovered that a current-carrying wire deflects a compass needle, proving moving charges produce magnetic fields.
  2. The Lorentz force on a charged particle is F = q(E + v × B).
  3. Magnetic force is always perpendicular to velocity, so it does no work and does not change kinetic energy.
  4. Biot-Savart law gives the magnetic field due to a current element: dB = (μ₀/4π) (I dl × r̂)/r².
  5. The magnetic field at the centre of a circular loop is B = μ₀I/(2R).
  6. Ampere's circuital law: ∮ B·dl = μ₀ I_enclosed.
  7. The magnetic field due to a long straight wire is B = μ₀I/(2πr).
  8. Inside a long solenoid, B = μ₀ n I, and outside it is nearly zero.
  9. Parallel currents attract; antiparallel currents repel.
  10. The force per unit length between two parallel wires is f = μ₀ I₁ I₂/(2π d).
  11. The ampere is defined as the current that produces 2 × 10⁻⁷ N/m between two parallel wires 1 m apart.
  12. Torque on a current loop: τ = I A B sin θ, or τ = m × B.
  13. The cyclotron frequency ω = qB/m is independent of particle speed.
  14. A charged particle moving at an angle to a uniform magnetic field follows a helical path.
  15. The pitch of a helix is p = v∥ T = 2π m v∥/(qB).
  16. The radius of circular motion in a magnetic field is r = mv/(qB).
  17. The magnetic field inside a current-carrying wire (r < a) is B = (μ₀ I r)/(2π a²).
  18. The magnetic field lines form closed loops, unlike electric field lines which start and end on charges.
  19. The permeability of free space μ₀ = 4π × 10⁻⁷ T m/A.
  20. The unit of magnetic field is tesla (T); 1 T = 1 N/(A·m); 1 G = 10⁻⁴ T.

MCQs — 10 Questions with Answers

Q1. A magnetic compass needle placed near a long straight current-carrying wire experiences deflection. According to Oersted's observation, the magnetic field lines around the wire are arranged as:

  • A. Concentric circles with the wire as centre, in a plane perpendicular to the wire ✓
  • B. Radial lines emerging directly outward from the wire in all directions
  • C. Parallel lines running along the length of the wire
  • D. Elliptical paths converging at the wire ends

Answer: A — Oersted's experiments showed that compass needles align tangentially to imaginary circles surrounding the wire; iron filings confirm this concentric circular field pattern.

Q2. An electron (charge −e) moves with velocity v perpendicular to a uniform magnetic field B. If a proton (charge +e) moves with the same velocity v in the same field, how do the magnitudes of magnetic forces compare?

  • A. Proton force is twice the electron force
  • B. Both experience equal magnitude of force, but in opposite directions ✓
  • C. Electron force is twice the proton force
  • D. Forces depend on mass, so they are different

Answer: B — Magnetic force magnitude is F = |q|vB sin(90°) = |q|vB; both have equal charge magnitude |e|, so |F| is the same; opposite signs give opposite force directions.

Q3. Which of the following statements about the magnetic force on a moving charge is correct?

  • A. The magnetic force does positive work, increasing the kinetic energy of the charge
  • B. The magnetic force is always perpendicular to the velocity, so it does zero work on the charge ✓
  • C. The magnetic force is parallel to the velocity and opposes motion
  • D. The magnetic force increases the speed of the charge by accelerating it along the direction of motion

Answer: B — The cross product v × B produces a force perpendicular to v; since work W = F·s requires component of force along displacement, perpendicular force does zero work.

Q4. A charged particle moving with velocity v at angle 30° to a magnetic field B experiences a magnetic force. If the velocity is changed to make angle 90° with B (keeping |v| and |B| constant), how does the force magnitude change?

  • A. Decreases by factor of 2
  • B. Increases by factor of 2 ✓
  • C. Remains the same
  • D. Increases by factor of √3

Answer: B — F = qvB sin θ; at 30°, F₁ = qvB sin(30°) = qvB(1/2); at 90°, F₂ = qvB sin(90°) = qvB; ratio F₂/F₁ = 2.

Q5. Which of the following is NOT correct about the Lorentz force on a moving charge in combined electric and magnetic fields?

  • A. The magnetic component of force depends on the cross product v × B
  • B. The electric component qE is independent of the velocity of the charge
  • C. Both electric and magnetic forces can change the speed of the particle ✓
  • D. The direction of force on a negative charge is opposite to that on a positive charge moving with the same velocity

Answer: C — Electric force can change speed (work can be done), but magnetic force is always perpendicular to v and does zero work; only electric force changes speed.

Q6. In the SI system, the unit of magnetic field (tesla) can be expressed as:

  • A. kg/(A·s²)
  • B. N·s/(C·m)
  • C. J/(A·m²)
  • D. All of the above are equivalent expressions ✓

Answer: D — From F = qvB, [B] = N/(C·m/s) = N·s/(C·m); N = kg·m/s², so T = kg/(A·s²); also J = N·m, so T = J/(A·m²) — all three are dimensionally equivalent.

Q7. A proton and an alpha particle (both positively charged) enter a uniform magnetic field with the same velocity perpendicular to the field. The alpha particle has charge 2e and mass 4m_p (where m_p is proton mass, e is elementary charge). Compare the magnitudes of magnetic forces on them:

  • A. Force on alpha particle is 2 times force on proton ✓
  • B. Force on proton is 2 times force on alpha particle
  • C. Both experience equal magnetic force
  • D. Force on alpha particle is 4 times force on proton

Answer: A — Magnetic force F = qvB; for proton: F_p = evB; for alpha: F_α = 2evB; ratio F_α/F_p = 2 (magnetic force depends on charge, not mass).

Q8. When current direction in a long straight wire is reversed, what happens to the magnetic field pattern around the wire?

  • A. The field magnitude doubles but direction remains the same
  • B. The field magnitude halves
  • C. The field direction reverses; compass needles flip to opposite orientation ✓
  • D. The field pattern transforms from circular to radial

Answer: C — Reversing current reverses the sense of circulation of the magnetic field around the wire; compass needles (which align with field) flip to opposite directions.

Q9. Assertion (A): The magnetic force on a moving charge always acts in a direction perpendicular to its velocity. Reason (R): This is because the magnetic force arises from the cross product v × B, which is always perpendicular to both v and B. Choose the correct option:

  • A. Both A and R are true, and R is the correct explanation of A ✓
  • B. Both A and R are true, but R is not the correct explanation of A
  • C. A is true but R is false
  • D. Both A and R are false

Answer: A — Assertion is correct: F = q(v × B) is always ⊥ to v. Reason is correct: by definition of cross product, v × B ⊥ to both v and B. R explains A perfectly.

Q10. A charged particle with charge q = 2 μC, moving with speed v = 10⁶ m/s perpendicular to a magnetic field B = 0.5 T, experiences a magnetic force. Calculate the magnitude of this force. (μ = 10⁻⁶)

  • A. 1 N ✓
  • B. 1 mN
  • C. 10 N
  • D. 0.1 N

Answer: A — F = qvB sin(90°) = qvB = (2 × 10⁻⁶ C)(10⁶ m/s)(0.5 T) = 2 × 10⁶ × 0.5 × 10⁻⁶ = 1 N.

Flashcards

State Oersted's observation about current and magnetism.

A current-carrying wire produces a magnetic field in concentric circles around the wire, which deflects a compass needle aligned tangentially.

Write the Lorentz force equation for a moving charge in electric and magnetic fields.

F = q[E(r) + v × B(r)], where the magnetic force is qv × B perpendicular to both velocity and field.

Why does a stationary charge not experience a magnetic force?

The magnetic force depends on velocity (v); when v = 0, the force qv × B = 0 regardless of B magnitude.

Define the unit of magnetic field (tesla) using the Lorentz force equation.

1 tesla is the magnetic field strength that exerts 1 newton force on a 1 coulomb charge moving at 1 m/s perpendicular to the field.

What does the right-hand rule tell us about the magnetic force direction?

Point fingers in direction of v, curl them toward B, thumb points in direction of v × B, giving the force direction on a positive charge.

How does reversing current direction affect the magnetic field pattern around a wire?

Reversing current reverses the direction of the magnetic field; compass needles flip to point in the opposite tangential direction (opposite circulation).

Why is the magnetic force zero when velocity is parallel or anti-parallel to the magnetic field?

The cross product v × B = 0 when the angle between v and B is 0° or 180°, because sin(0°) = sin(180°) = 0.

State the principle of superposition for magnetic fields.

The total magnetic field from multiple sources is the vector sum of individual magnetic fields from each source.

What is the relationship between magnetic force and work done on a charge?

Magnetic force does zero work because it is always perpendicular to velocity; only the electric component can do work.

Name the physical discovery that unified electricity and magnetism, and who formulated it.

James Maxwell unified electricity and magnetism in 1864 by formulating Maxwell's equations, which also predicted light as electromagnetic waves.

Important Board Questions

Define magnetic field B and explain why a stationary charge does not experience a magnetic force, even though a moving charge does. [2 marks]

Magnetic field is a vector field created by moving charges; use Lorentz force equation F = q(v × B) and explain that when v = 0, cross product becomes zero.

A charged particle of mass m and charge q enters a uniform magnetic field B with velocity v perpendicular to the field. Using the right-hand rule, explain the direction of the magnetic force and show that this force does zero work on the particle. What can you deduce about the particle's speed? [5 marks]

Apply right-hand rule for v × B to find force direction (perpendicular to both v and B); use work definition W = F·s cos θ where θ = 90°; conclude that magnetic force cannot change kinetic energy or speed, only direction.

Describe Oersted's experiment that demonstrated the connection between electric current and magnetism. Explain how the arrangement of iron filings around a current-carrying wire confirms the pattern of the magnetic field, and state the principle of superposition for magnetic fields. How did this discovery lead to Maxwell's unification of electricity and magnetism? [6 marks]

Describe compass needle deflection in tangential pattern, iron filings arranging in concentric circles, field reversal with current reversal; explain superposition as vector addition of B fields from multiple sources; link to historical development: Oersted (1820) → Maxwell (1864) unified laws → predicted light as EM waves.

Assertion-Reason Questions for Moving Charges and Magnetism

Q1. Assertion (A):

A moving charged particle in a magnetic field experiences a force that does no work on the particle.

Reason (R):

The magnetic force is always perpendicular to the velocity of the charged particle.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The magnetic force F = q(v × B) is by definition perpendicular to v (due to the cross product). Since work W = F·d = Fd cos(90°) = 0 when force is perpendicular to displacement, the magnetic force does no work. The reason correctly explains why the assertion is true.

Q2. Assertion (A):

When a compass needle is placed near a current-carrying wire, the needle always aligns along the direction of the magnetic field produced by the current.

Reason (R):

The compass needle aligns perpendicular to the wire carrying the current.

Answer: C — A is true but R is false — The assertion is true—the needle aligns along the magnetic field lines. However, the reason is false. The needle does not align perpendicular to the wire; it aligns tangentially to the circular field lines that encircle the wire. The reason incorrectly describes the geometry.

Q3. Assertion (A):

The Lorentz force on a charged particle is the vector sum of electric and magnetic forces.

Reason (R):

The electric force depends on both the charge and the electric field, while the magnetic force depends on charge, velocity, and magnetic field.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both assertion and reason are true. F = q[E + (v × B)] is indeed the vector sum. The reason correctly identifies the dependences of both forces. However, the reason does not explain why it is a vector sum (that would involve discussing superposition principle), so it is not the correct explanation of the assertion.

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Case-Based Questions (CBQ) for Moving Charges and Magnetism

Case Study:

A research team at TIFR (Tata Institute of Fundamental Research) is studying the behaviour of cosmic ray protons entering Earth's magnetosphere. They observe that a 2 GeV proton enters Earth's magnetic field region perpendicularly at the equator, where the magnetic field strength is approximately 3 × 10⁻⁵ T. The proton follows a circular path with a radius of about 7000 km, which is comparable to Earth's radius. The team wants to understand how the proton's trajectory changes if it enters at different speeds or locations with different field strengths.

Q1. The radius of the proton's circular path in Earth's magnetic field is given by r = mv/(qB). If the proton's speed is doubled while keeping B constant, what happens to the radius?

  • A. Radius becomes half
  • B. Radius doubles ✓
  • C. Radius becomes one-quarter
  • D. Radius remains unchanged

Answer: B

Q2. At a location where Earth's magnetic field is twice as strong (6 × 10⁻⁵ T), what would be the radius of the path for a proton with the same speed as before?

  • A. 14000 km
  • B. 7000 km
  • C. 3500 km ✓
  • D. 1750 km

Answer: C

Q3. Using the right-hand rule, if a proton enters moving northward and Earth's magnetic field at the equator points northward, in which direction will the proton experience the magnetic force?

  • A. Northward (parallel to the field)
  • B. Downward (toward Earth's centre)
  • C. Horizontal (perpendicular to both velocity and field)
  • D. Zero force (velocity parallel to field) ✓

Answer: D

Q4. If an alpha particle (charge +2e, mass ≈ 4u) and a proton (charge +e, mass ≈ u) both enter the same magnetic field region with the same kinetic energy, which follows the tighter circular path?

  • A. The alpha particle, because it has greater charge
  • B. The proton, because it has smaller mass
  • C. The alpha particle, because its radius depends inversely on charge ✓
  • D. They follow identical paths since kinetic energy is equal

Answer: C

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Previous Year Questions (PYQ) for Moving Charges and Magnetism

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance (a/2) from the axis of the wire will be

  • Zero
  • μ₀I/2πa
  • μ₀I/4πa
  • μ₀I/6πa

CBSE 2026 · Set 55/1/1 · 3 marks

Figure shows a narrow beam of electrons entering with a velocity of 3 × 10⁷ m/s, symmetrically through the space between two parallel horizontal plates P1 P1' and P2 P2' kept 2 cm apart. If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end P2'.

CBSE 2026 · Set 55/1/2 · 1 mark

A straight conductor lies along x-axis. It carries a current of 10 A along +x direction. The magnetic field B⃗ due to 1 cm segment of this conductor, centred at the origin, at a point (0, 1 m, 0) is

  • (1 nT) ĵ
  • (10 nT) k̂
  • − (10 nT) k̂
  • − (1 nT) ĵ

CBSE 2026 · Set 55/1/2 · 2 marks

A wire of length L is bent round into (i) a square coil having N turns and (ii) a circular coil having N turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.

CBSE 2026 · Set 55/1/2 · 4 marks

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances.

The torque on the coil remains constant irrespective of the coil's orientation during rotation due to

  • use of soft iron core which increases the magnetic field.
  • radial magnetic field
  • hair spring which provides the counter torque
  • eddy current in the iron core which causes damping.

CBSE 2026 · Set 55/1/3 · 1 mark

A circular loop has radius R and carries current I as shown in figure. In order that the net magnetic field at the centre of the loop is zero, the current in wire AB should have magnitude

  • 2 I, along +X – axis
  • 2 I along –X – axis
  • I along +X – axis
  • I along –X – axis

CBSE 2026 · Set 55/1/3 · 4 marks

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances.

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances. (I) The torque on the coil remains constant irrespective of the coil's orientation during rotation due to

  • use of soft iron core which increases the magnetic field.
  • radial magnetic field
  • hair spring which provides the counter torque
  • eddy current in the iron core which causes damping.

CBSE 2026 · Set 55/2/1 · 1 mark

A straight long wire lying along y-axis, carries a current of 1 A along −y direction. The magnetic field due to the conductor at a point (50 cm, 0, 0) will point along.

  • z-axis
  • −z-axis
  • x-axis
  • −x-axis

CBSE 2026 · Set 55/2/1 · 3 marks

Two long straight parallel conductors A and B carrying steady currents Iₐ and I_b in the same direction are separated by a distance d. Deduce the expressions for the force acting on length L of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length L of conductor A due to conductor B and show that it follows Newton's third law.

CBSE 2026 · Set 55/2/1 · 3 marks

Derive an expression for the magnetic field B⃗, due to a circular coil of N turns, each of radius r carrying current I, at a distance 'x' from the centre along its axis.

CBSE 2026 · Set 55/2/2 · 1 mark

A straight conductor lies along x-axis and carries a current of 2 A along +x direction. The magnetic field at a point (0, 40 cm, 0) due to 1 cm length of conductor centered at the origin points along

  • y-axis
  • −y-axis
  • z-axis
  • −z-axis

CBSE 2026 · Set 55/2/2 · 3 marks

Two long straight parallel conductors A and B carrying steady currents Iₐ and Iᵦ in the same direction are separated by a distance d. Deduce the expressions for the force acting on length L of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length L of conductor A due to conductor B and show that it follows Newton's third law.

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Class 12 Physics Formula Sheet

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Class 12 Physics Previous Year Questions, Chapter-wise

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Class 12 Physics Derivations

All 68 derivations, chapter-wise, step by step with diagrams and marked with the CBSE board years (2022–2026) each was asked in. Free PDF download.

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