← Physics (312) · Class 12

L-18: Magnetism and Magnetic Effect of Electric Current

Physics — Class 12 · NIOS Code 312 · Module 5 · Source: 312_Physics_Eng_Lesson18.pdf

Magnetism and Magnetic Effect of Electric Current

Oersted discovered the intimate link between electricity and magnetism. This lesson covers magnets, earth's magnetism, magnetic fields due to currents, forces on moving charges and conductors, cyclotrons, and measuring instruments.

NIOS objectives: magnetic field; earth's field elements; Oersted's experiment; Biot-Savart and Ampere's laws; charged particle motion; cyclotron; force on current conductor; parallel wire force; galvanometer, ammeter, voltmeter.

18.1 Magnets and Their Properties

  • Directive property: freely suspended magnet aligns ~N–S (geographic).
  • Attractive property: attracts iron, nickel, cobalt; poles at ends.
  • Like poles repel; unlike poles attract.
  • Poles are inseparable — simplest magnetic source is a dipole.
  • Magnetic induction: iron near a magnet acquires induced polarity.

18.1.1 Magnetic Field Lines

  • Tangent to a line gives direction of B.
  • Density of lines ∝ field strength.
  • Outside magnet: N → S; inside: S → N (closed curves).
  • Two field lines never cross.
Fig 18.3 — Bar Magnet Field Lines N S closed curves: N→S outside, S→N inside
Fig 18.3 — Magnetic field lines form closed loops through a bar magnet

Earth's Magnetic Field

Earth acts as a magnet; magnetic S pole near geographic N. Magnetic axis MM₁ does not coincide with rotation axis RR₁. Field magnitude and direction change with time.

Three elements:

  • Dip (inclination) δ: angle between B and horizontal in magnetic meridian.
  • Declination θ: angle between magnetic and geographic meridians.
  • Horizontal component BH: component of B along horizontal.
BH = B cos δ  |  BV = B sin δ
B = total earth's field at a point
BH² + BV² = B²
BV/BH = tan δ

18.2 Electricity and Magnetism — Oersted's Experiment

A current-carrying conductor produces a magnetic field around it (Oersted, 1820). Compass needle deflects near a current-carrying wire; deflection reverses when current reverses. Field lines around a straight wire are concentric circles.

Fig 18.7 — Field Around Current-Carrying Wire ⊙ I into page B (circular) N Right-hand rule: thumb = I, curled fingers = B direction
Fig 18.7 — Concentric magnetic field lines around a straight current-carrying conductor

18.3 Biot-Savart's Law

Each current element Δl contributes to B; net field is vector sum:

|dB| = (μ₀/4π) × (I dl sin θ) / r²
μ₀ = 4π×10⁻⁷ T·m·A⁻¹ (permeability of vacuum)
θ = angle between dl and line to point P
r = distance from element to P
In medium: B = μ₀μᵣ B_vacuum

Right-hand grip rule: thumb along current, curled fingers show B direction.

18.3.1 Magnetic Field at Centre of Circular Coil

B = μ₀ N I / (2r)
N = number of turns
r = radius of coil
Field perpendicular to plane of coil
End rule: clockwise current → S-pole face; anticlockwise → N-pole face

18.4 Ampere's Circuital Law

∮ B · dl = μ₀ I
Line integral of B around any closed loop = μ₀ × enclosed current
Independent of loop size/shape (for symmetric cases)
Complements Biot-Savart for simple geometries

Applications

Long straight wire: B = μ₀ I / (2πr)
Circular Amperian loop concentric with wire
Field falls as 1/r
Solenoid (inside): B = μ₀ n I
n = turns per unit length
Uniform field along axis near centre
At ends: B = μ₀ n I / 2; outside ≈ 0
Toroid: B = μ₀ N I / (2πr)
Endless solenoid bent into a circle
Field only inside core; zero outside

Electromagnet: current-carrying solenoid with soft iron core. Strength ∝ n and I. Displacement current (Maxwell): time-varying electric field also produces magnetic field — completes Ampere's law for capacitors.

18.5 Force on Moving Charge and Current

Lorentz Force

F = q (v × B)  |  |F| = q v B sin θ
Fleming's left-hand rule: forefinger = B, middle finger = v (for +q), thumb = F
No work done by magnetic force (F ⊥ v) — speed unchanged
Max when θ = 90°: F = qvB
Fig 18.17 — Fleming's Left-Hand Rule B (field) forefinger v (motion) middle finger F (force) thumb B I F
Fig 18.17 — Fleming's left-hand rule gives direction of force on current/charge in B

18.5.1 Force on Current-Carrying Conductor

F = B I L sin θ
L = length of conductor in field
Unit of B: tesla (T) = N·A⁻¹·m⁻¹
1 T defined from F = BIL when F, I, L mutually perpendicular

18.5.2 Force Between Parallel Wires

F/l = μ₀ I₁ I₂ / (2πr)
r = separation between wires
Same direction currents → attract
Opposite directions → repel
Definition of 1 A: force 2×10⁻⁷ N/m between 1 A wires 1 m apart

18.5.3 Motion in Uniform Magnetic Field

R = m v / (q B)  |  T = 2πm / (q B)
Perpendicular entry → circular path
R ∝ momentum; R ∝ 1/B
Time period T independent of v and R
θ ≠ 0°, 90° → helical path; θ = 0° → straight line

18.5.5 Cyclotron

Device (Lawrence, 1929) to accelerate charged particles using dees (D-shaped electrodes) in perpendicular magnetic field and oscillating electric field in the gap.

vc = q B / (2πm)
Cyclotron frequency independent of radius and speed
When oscillator frequency = vc, particle gains energy each crossing (resonance)
Maximum energy limited by dee radius

18.6 Current Loop as a Magnetic Dipole

M = N I A
M = magnetic dipole moment (A·m²)
Current loop ≡ bar magnet with N and S faces
Axial field: B = μ₀M/(2πx³) at far point
Equatorial: B = −μ₀M/(4πx³)
τ = M × B = N B I A sin θ
No net force in uniform B, only torque
Tends to align M with B
Basis of electric motors and galvanometers

Magnetism in Matter

  • Diamagnetic: feebly repelled (bismuth, copper).
  • Paramagnetic: feebly attracted (aluminium).
  • Ferromagnetic: strongly attracted (Fe, Ni, Co) — domains align in external field; Curie temperature Tc above which paramagnetic.

18.6.2 Galvanometer

Moving-coil instrument: current in coil in radial magnetic field (curved pole pieces + soft iron core) produces torque; spring provides restoring torque.

I = (k / N B A) × α
α = deflection angle ∝ current
k = torsional constant of spring
Sensitivity ↑ with large N, B, A and small k
Detects currents ~0.1 μA

18.6.3 Ammeter and Voltmeter

Ammeter shunt: S = Ig G / (I − Ig)
Low resistance in parallel with galvanometer
Very low effective resistance — connected in series
Ideal ammeter: zero resistance
Voltmeter: R = V/Ig − G
High resistance in series with galvanometer
Connected in parallel across points to measure V
Ideal voltmeter: infinite resistance
Fig 18.27 — Moving Coil Galvanometer N S coil core pointer scale radial B + soft iron core → τ ∝ I → α ∝ I
Fig 18.27 — Galvanometer: coil in radial field; deflection proportional to current
         MAGNETISM — KEY FORMULAS
         ========================
    Biot-Savart     :  dB = (μ₀/4π) I dl sinθ / r²
    Coil centre     :  B = μ₀NI / (2r)
    Long wire       :  B = μ₀I / (2πr)
    Solenoid        :  B = μ₀nI
    Ampere's law    :  ∮B·dl = μ₀I
    Lorentz force   :  F = q(v × B)
    Conductor       :  F = BIL sinθ
    Parallel wires  :  F/l = μ₀I₁I₂ / (2πr)
    Circular motion :  R = mv/(qB),  T = 2πm/(qB)
    Dipole moment   :  M = NIA
    Torque          :  τ = M × B
    Cyclotron freq  :  νc = qB/(2πm)

Quick Revision

  • Magnet poles inseparable; field lines closed N→S outside.
  • Earth: dip δ, declination θ, BH = B cos δ.
  • Current → magnetic field (Oersted); μ₀ = 4π×10⁻⁷ T·m·A⁻¹.
  • Biot-Savart and Ampere's law for B due to currents.
  • F = qvB (Lorentz); F = BIL on conductor; Fleming's left-hand rule.
  • Parallel wires: attract if currents same direction.
  • Charged particle: circular path R = mv/qB; cyclotron vc = qB/2πm.
  • M = NIA; τ = MB sin θ; galvanometer → ammeter (shunt) / voltmeter (series R).
20 cards · click any card to flip
Magnetic dipole
Simplest magnetic source — poles are inseparable. A bar magnet or current loop (M = NIA) acts as a dipole.
Magnetic field lines
Tangent = B direction. Density ∝ field strength. Never cross. Outside magnet N→S; inside S→N (closed loops).
Earth's magnetic elements
Dip (inclination) δ, declination θ, horizontal component B_H = B cos δ. Magnetic axis ≠ geographic axis.
Oersted's discovery (1820)
Electric current produces a magnetic field around the conductor. Compass deflects near current-carrying wire.
μ₀ (permeability of vacuum)
μ₀ = 4π × 10⁻⁷ T·m·A⁻¹ (or Wb·A⁻¹·m⁻¹). Appears in Biot-Savart and Ampere's law.
Biot-Savart law
dB = (μ₀/4π) I dl sinθ / r². Gives B from a current element. Right-hand rule for direction.
Field at centre of circular coil
B = μ₀NI/(2r). End rule: clockwise current → S face; anticlockwise → N face.
Ampere's circuital law
∮B·dl = μ₀I (enclosed current). Used for long wire, solenoid, toroid. Long wire: B = μ₀I/(2πr).
Solenoid field
Inside (centre): B = μ₀nI. At ends: μ₀nI/2. Outside ≈ 0. n = turns per unit length. Behaves like bar magnet.
Lorentz force
F = q(v × B); |F| = qvB sinθ. Fleming's left-hand rule. No work done — speed unchanged in pure B field.
Force on current conductor
F = BIL sinθ. Defines tesla: 1 T = 1 N·A⁻¹·m⁻¹. Maximum when conductor ⊥ B.
Parallel current-carrying wires
F/l = μ₀I₁I₂/(2πr). Same direction → attract. Opposite → repel. Basis of ampere definition.
Charged particle in B field
Perpendicular v → circle R = mv/(qB). T = 2πm/(qB) independent of v. Oblique → helix.
Cyclotron
Accelerates ions using dees + perpendicular B + oscillating E. νc = qB/(2πm). Energy limited by dee radius.
Magnetic dipole moment
M = NIA for a current loop. Analogous to electric dipole. τ = M × B in uniform field.
Torque on current coil
τ = NBIA sinθ = MB sinθ. No net force in uniform B. Principle of motors and galvanometers.
Ferromagnetism
Fe, Ni, Co strongly attracted. Domains align in external field. Above Curie temperature Tc → paramagnetic.
Galvanometer
Detects current via coil torque in radial B field. α ∝ I. Soft iron core increases sensitivity.
Ammeter conversion
Low resistance shunt S in parallel: S = IgG/(I−Ig). Very low resistance; connected in series in circuit.
Voltmeter conversion
High resistance R in series: R = V/Ig − G. High resistance; connected in parallel across points.

Q1. The SI unit of magnetic field is:

Q2. Oersted's experiment demonstrated that:

Q3. Magnetic field at distance r from a long straight wire carrying current I is:

Q4. Magnetic field inside a long solenoid near its centre is:

Q5. Lorentz force on a charge q moving with velocity v in magnetic field B is:

Q6. Force on a conductor of length L carrying current I perpendicular to uniform B is:

Q7. Two parallel wires carrying currents in the same direction:

Q8. Radius of circular path of charge q with speed v perpendicular to B is:

Q9. Magnetic dipole moment of a coil with N turns, current I, area A is:

Q10. To convert a galvanometer into an ammeter, we connect:

B_H = B cos δ  |  B_V = B sin δ
|dB| = (μ₀/4π)(I dl sin θ)/r²
B = μ₀ N I / (2r)
∮ B · dl = μ₀ I
B = μ₀ I / (2πr)
B = μ₀ n I
B = μ₀ N I / (2πr)
F = q(v × B)  |  |F| = qvB sin θ
F = B I L sin θ
F/l = μ₀ I₁ I₂ / (2πr)
R = mv/(qB)  |  T = 2πm/(qB)
ν_c = qB / (2πm)
M = N I A
τ = M × B = N B I A sin θ
I = (k / N B A) × α
S = I_g G / (I − I_g)
R = V/I_g − G

1. Formulas & Definitions

Full Ch 18 study guide — Earth's field, Biot-Savart, Ampere's law, Lorentz force, cyclotron, dipoles, and measuring instruments.

B_H = B cos δ  |  B_V = B sin δ

Definition: Horizontal and vertical components of Earth's magnetic field.

Derivation

Resolve total field B at angle δ (dip/inclination) from horizontal.

Variables

B_H, B_V (T) · δ = dip angle · B_V/B_H = tan δ

Why it works

Navigation and compass readings use B_H; dip angle varies with latitude.

Historical context

Earth's magnetic S pole near geographic N; declination θ is separate (meridian angle).

Deep understanding

B_H² + B_V² = B². At equator δ≈0; at poles δ≈90°.

2. Diagrams & Visuals

δ B_H

Color-coded visual · step-by-step breakdown below

  1. Know total B and dip δ
  2. B_H = B cos δ
  3. B_V = B sin δ
  4. Check tan δ = B_V/B_H

3. Solved Examples

Basic

Q: B=0.4 G, δ=30°.

Solution: B_H≈0.35 G

Answer: ~0.35 G

Intermediate

Q: δ=0° at equator?

Solution: B_H=B

Answer: Full horizontal

Advanced

Q: δ=90°?

Solution: B_H=0

Answer: Vertical only

Exam

Q: Earth field horizontal?

Solution: B cos δ

Answer: Sec 18.1

|dB| = (μ₀/4π)(I dl sin θ)/r²

Definition: Biot-Savart law — magnetic field due to current element Idl.

Derivation

Each element contributes dB; net B is vector sum over wire.

Variables

μ₀ = 4π×10⁻⁷ T·m/A⁻¹ · θ = angle dl to line to P · r = distance

Why it works

Fundamental law for B from arbitrary current distributions.

Historical context

Oersted (1820) linked current and magnetism; Biot-Savart gives quantitative field.

Deep understanding

In medium: B = μ₀μᵣ B_vac. Right-hand grip rule for direction.

2. Diagrams & Visuals

dB ∝ Idl sinθ/r²

Color-coded visual · step-by-step breakdown below

  1. Identify current element Idl
  2. Angle θ and distance r to P
  3. dB magnitude from formula
  4. Vector sum all elements

3. Solved Examples

Basic

Q: θ=90°, double r?

Solution: dB quarters

Answer: 1/r²

Intermediate

Q: θ=0° (on axis)?

Solution: sinθ=0, dB=0

Answer: Zero contribution

Advanced

Q: μ₀ value?

Solution: 4π×10⁻⁷

Answer: SI constant

Exam

Q: Biot-Savart?

Solution: μ₀/4π · Idl sinθ/r²

Answer: Sec 18.3

B = μ₀ N I / (2r)

Definition: Magnetic field at centre of circular coil of N turns, radius r.

Derivation

Integrate Biot-Savart around circle; all elements contribute in same direction at centre.

Variables

N = turns · r = radius (m) · B perpendicular to coil plane

Why it works

Used in galvanometers, electromagnets, and field measurement.

Historical context

End rule: clockwise current → S face; anticlockwise → N face.

Deep understanding

Doubling N or I doubles B; doubling r halves B.

2. Diagrams & Visuals

B = μ₀NI/2r

Color-coded visual · step-by-step breakdown below

  1. Count turns N, radius r
  2. Current I through coil
  3. B = μ₀NI/(2r)
  4. Direction via grip/end rule

3. Solved Examples

Basic

Q: N=100, I=2 A, r=0.1 m.

Solution: B≈1.26 mT

Answer: ~1.26 mT

Intermediate

Q: Double N?

Solution: B doubles

Answer: 2B

Advanced

Q: Double r?

Solution: B halves

Answer: B ∝ 1/r

Exam

Q: Coil centre field?

Solution: μ₀NI/2r

Answer: Sec 18.3.1

∮ B · dl = μ₀ I

Definition: Ampere's circuital law — line integral of B around closed loop.

Derivation

Relates circulation of B to current enclosed by Amperian loop.

Variables

I = net current through loop (signed) · μ₀ permeability

Why it works

Simpler than Biot-Savart for symmetric geometries (wire, solenoid, toroid).

Historical context

Maxwell added displacement current term for complete theory.

Deep understanding

Choose Amperian loop matching symmetry; B constant on loop segment.

2. Diagrams & Visuals

∮B·dl = μ₀I

Color-coded visual · step-by-step breakdown below

  1. Draw closed Amperian loop
  2. Sum B·dl around loop
  3. Equate to μ₀ × enclosed I
  4. Solve for B

3. Solved Examples

Basic

Q: Symmetric loop, B uniform.

Solution: B×2πr = μ₀I

Answer: Standard wire

Intermediate

Q: No enclosed current?

Solution: ∮B·dl=0

Answer: Outside toroid

Advanced

Q: Multiple wires inside?

Solution: Use net I

Answer: Superposition

Exam

Q: Ampere's law?

Solution: ∮B·dl=μ₀I

Answer: Sec 18.4

B = μ₀ I / (2πr)

Definition: Magnetic field at distance r from long straight current-carrying wire.

Derivation

Circular Amperian loop concentric with wire; B tangential and constant on loop.

Variables

r = perpendicular distance from wire (m) · I (A)

Why it works

Classic Oersted result; field lines are concentric circles.

Historical context

B falls as 1/r. Right-hand rule: thumb = I, fingers curl = B.

Deep understanding

Field same magnitude at all points on circle of radius r.

2. Diagrams & Visuals

B = μ₀I/2πr

Color-coded visual · step-by-step breakdown below

  1. Distance r from wire
  2. Current I
  3. B = μ₀I/(2πr)
  4. Direction: right-hand rule

3. Solved Examples

Basic

Q: I=10 A, r=0.05 m.

Solution: B=4×10⁻⁵ T

Answer: 40 μT

Intermediate

Q: Double r?

Solution: B halves

Answer: 1/r

Advanced

Q: Reverse current?

Solution: B direction reverses

Answer: Sign of I

Exam

Q: Long wire field?

Solution: μ₀I/2πr

Answer: Sec 18.4

B = μ₀ n I

Definition: Magnetic field inside long solenoid near centre.

Derivation

Rectangular Amperian loop: outer segments B≈0; inner segment B×l = μ₀(nl)I.

Variables

n = turns per unit length (m⁻¹) · I = current

Why it works

Uniform strong field for electromagnets, MRI coils, inductors.

Historical context

At ends B = μ₀nI/2; outside ≈ 0 for ideal long solenoid.

Deep understanding

Insert soft iron core → μᵣ increases B further (electromagnet).

2. Diagrams & Visuals

B = μ₀nI

Color-coded visual · step-by-step breakdown below

  1. Turns per length n
  2. Current I
  3. B = μ₀nI inside
  4. Weaker at ends

3. Solved Examples

Basic

Q: n=1000/m, I=2 A.

Solution: B≈2.5 mT

Answer: ~2.5 mT

Intermediate

Q: Double n?

Solution: B doubles

Answer: 2B

Advanced

Q: Outside ideal solenoid?

Solution: B≈0

Answer: Negligible

Exam

Q: Solenoid centre B?

Solution: μ₀nI

Answer: Sec 18.4

B = μ₀ N I / (2πr)

Definition: Magnetic field inside toroid (endless solenoid bent in circle).

Derivation

Amperian circle inside core: B×2πr = μ₀NI.

Variables

N = total turns · r = radius of toroid cross-section centre

Why it works

Confines field inside ring — no external stray field.

Historical context

Toroid outside B ≈ 0; all flux closed in core.

Deep understanding

Unlike solenoid, field zero outside toroid completely.

2. Diagrams & Visuals

B = μ₀NI/2πr

Color-coded visual · step-by-step breakdown below

  1. Total turns N on toroid
  2. Mean radius r
  3. B = μ₀NI/(2πr)
  4. Field only in core

3. Solved Examples

Basic

Q: N=500, I=1 A, r=0.1 m.

Solution: B≈1 mT

Answer: ~1 mT

Intermediate

Q: Outside toroid?

Solution: B≈0

Answer: Confined flux

Advanced

Q: vs solenoid formula?

Solution: Uses 2πr path

Answer: Toroidal geometry

Exam

Q: Toroid field?

Solution: μ₀NI/2πr

Answer: Sec 18.4

F = q(v × B)  |  |F| = qvB sin θ

Definition: Lorentz force on moving charge in magnetic field.

Derivation

Force perpendicular to both v and B; magnitude max when θ=90°.

Variables

q (C) · v (m/s) · B (T) · θ between v and B

Why it works

Explains deflection of charges, auroras, mass spectrometers.

Historical context

Fleming's left-hand rule: forefinger B, middle v (+q), thumb F.

Deep understanding

Magnetic force does no work (F⊥v) — speed unchanged, only direction.

2. Diagrams & Visuals

F = qvB sinθ

Color-coded visual · step-by-step breakdown below

  1. Identify q, v, B
  2. Angle θ between v and B
  3. |F| = qvB sinθ
  4. Direction: Fleming left-hand or v×B

3. Solved Examples

Basic

Q: q=e, v⊥B, B=1 T, v=10⁶.

Solution: F=1.6×10⁻¹³ N

Answer: Max force

Intermediate

Q: θ=0° parallel?

Solution: F=0

Answer: No deflection

Advanced

Q: Negative charge?

Solution: F direction opposite

Answer: Sign of q

Exam

Q: Lorentz force?

Solution: F=qvB sinθ

Answer: Sec 18.5

F = B I L sin θ

Definition: Force on straight current-carrying conductor in uniform magnetic field.

Derivation

N electrons → effective force F = (q/t)(L) × B = IL × B magnitude.

Variables

L = length in field (m) · θ between I and B · 1 T = N·A⁻¹·m⁻¹

Why it works

Basis of electric motors, loudspeakers, galvanometer torque.

Historical context

Tesla (T) defined from F = BIL when F, I, L mutually perpendicular.

Deep understanding

Maximum force when conductor ⊥ B (θ=90°).

2. Diagrams & Visuals

F = BIL sinθ

Color-coded visual · step-by-step breakdown below

  1. Current I, length L in field
  2. Angle θ between I and B
  3. F = BIL sinθ
  4. Fleming left-hand for direction

3. Solved Examples

Basic

Q: B=0.5 T, I=4 A, L=0.2 m, θ=90°.

Solution: F=0.4 N

Answer: 0.4 N

Intermediate

Q: θ=30°?

Solution: F halves vs 90°

Answer: sin30°=0.5

Advanced

Q: I parallel B?

Solution: F=0

Answer: θ=0

Exam

Q: Force on wire?

Solution: BIL sinθ

Answer: Sec 18.5.1

F/l = μ₀ I₁ I₂ / (2πr)

Definition: Force per unit length between two parallel current-carrying wires.

Derivation

Field from wire 1 acts on wire 2: F/L = I₂ × B₁.

Variables

r = separation (m) · I₁, I₂ (A)

Why it works

Defines the ampere; explains attraction/repulsion of parallel conductors.

Historical context

1 A official definition: 2×10⁻⁷ N/m between 1 A wires 1 m apart.

Deep understanding

Same direction currents attract; opposite directions repel.

2. Diagrams & Visuals

F/l ∝ I₁I₂/r

Color-coded visual · step-by-step breakdown below

  1. Currents I₁, I₂
  2. Separation r
  3. F/l = μ₀I₁I₂/(2πr)
  4. Same direction → attract

3. Solved Examples

Basic

Q: I₁=I₂=1 A, r=1 m.

Solution: F/l=2×10⁻⁷ N/m

Answer: Ampere def.

Intermediate

Q: Double r?

Solution: Force halves

Answer: 1/r

Advanced

Q: Opposite currents?

Solution: Repel

Answer: Not attract

Exam

Q: Parallel wire force?

Solution: μ₀I₁I₂/2πr

Answer: Sec 18.5.2

R = mv/(qB)  |  T = 2πm/(qB)

Definition: Radius and time period of circular motion of charge perpendicular to B.

Derivation

Magnetic force provides centripetal force: qvB = mv²/R.

Variables

R = radius · T = time period · m, q, v, B

Why it works

Mass spectrometers, bubble chambers, particle physics.

Historical context

T independent of v and R — key for cyclotron resonance.

Deep understanding

θ≠90° → helical path; θ=0° → straight line (no force).

2. Diagrams & Visuals

R = mv/qB

Color-coded visual · step-by-step breakdown below

  1. Charge enters ⊥ to B
  2. qvB = mv²/R → R = mv/qB
  3. T = 2πR/v = 2πm/qB
  4. Speed unchanged by B

3. Solved Examples

Basic

Q: Double v?

Solution: R doubles

Answer: R ∝ v

Intermediate

Q: Double B?

Solution: R halves

Answer: R ∝ 1/B

Advanced

Q: T depends on v?

Solution: No

Answer: T = 2πm/qB only

Exam

Q: Circular path radius?

Solution: mv/qB

Answer: Sec 18.5.3

ν_c = qB / (2πm)

Definition: Cyclotron frequency — angular frequency of particle in uniform B.

Derivation

From T = 2πm/qB; ν_c = 1/T = qB/(2πm).

Variables

ν_c (Hz) · independent of radius and speed

Why it works

Cyclotron resonance: oscillator at ν_c accelerates particle each gap crossing.

Historical context

Lawrence (1929); limited by relativistic mass increase and dee radius.

Deep understanding

When applied frequency = ν_c, particle gains energy synchronously.

2. Diagrams & Visuals

ν_c = qB/2πm

Color-coded visual · step-by-step breakdown below

  1. Charge q, mass m, field B
  2. ν_c = qB/(2πm)
  3. Match oscillator frequency
  4. Particle spirals outward

3. Solved Examples

Basic

Q: Proton in B=1 T.

Solution: ν_c≈15 MHz

Answer: ~15 MHz

Intermediate

Q: Double B?

Solution: ν_c doubles

Answer: 2ν_c

Advanced

Q: Depends on speed?

Solution: No (non-relativistic)

Answer: Constant ν_c

Exam

Q: Cyclotron frequency?

Solution: qB/2πm

Answer: Sec 18.5.5

M = N I A

Definition: Magnetic dipole moment of current loop.

Derivation

Loop of area A, N turns, current I behaves like bar magnet.

Variables

M (A·m²) · A = area of loop · direction ⊥ to loop (right-hand rule)

Why it works

Links current loops to bar magnets; basis for torque and motor action.

Historical context

Axial far field B = μ₀M/(2πx³); equatorial B = −μ₀M/(4πx³).

Deep understanding

Larger N, I, or A → stronger dipole.

2. Diagrams & Visuals

M = NIA

Color-coded visual · step-by-step breakdown below

  1. Area A of loop
  2. Turns N, current I
  3. M = NIA
  4. Direction via right-hand rule

3. Solved Examples

Basic

Q: N=50, I=0.5 A, A=0.01 m².

Solution: M=0.25 A·m²

Answer: 0.25 A·m²

Intermediate

Q: Double turns?

Solution: M doubles

Answer: 2M

Advanced

Q: Circular loop radius r?

Solution: A=πr²

Answer: Then M=NIπr²

Exam

Q: Dipole moment?

Solution: M=NIA

Answer: Sec 18.6

τ = M × B = N B I A sin θ

Definition: Torque on current loop (magnetic dipole) in uniform field B.

Derivation

Forces on opposite sides of loop create couple; τ = MB sin θ.

Variables

θ = angle between M and B · τ tends to align M with B

Why it works

Electric motors and galvanometers use this torque.

Historical context

No net force in uniform B — only rotation.

Deep understanding

Stable equilibrium when M ∥ B (θ=0); unstable when M ∥ −B.

2. Diagrams & Visuals

τ = M B sinθ

Color-coded visual · step-by-step breakdown below

  1. Dipole moment M = NIA
  2. Angle θ with B
  3. τ = MB sinθ
  4. Aligns M with B

3. Solved Examples

Basic

Q: M=0.1, B=0.5 T, θ=90°.

Solution: τ=0.05 N·m

Answer: 0.05 N·m

Intermediate

Q: θ=0°?

Solution: τ=0

Answer: Aligned

Advanced

Q: Max torque when?

Solution: θ=90°

Answer: sinθ=1

Exam

Q: Loop torque?

Solution: NBIA sinθ

Answer: Sec 18.6

I = (k / N B A) × α

Definition: Galvanometer — current proportional to coil deflection α.

Derivation

Torque NIAB = kα at equilibrium (spring restoring torque).

Variables

k = torsional constant · α = deflection angle · N, B, A of coil

Why it works

Sensitive current detection (~0.1 μA); basis for ammeter/voltmeter.

Historical context

Moving coil in radial magnetic field (curved poles + iron core).

Deep understanding

Sensitivity ↑ with large N, B, A and small k.

2. Diagrams & Visuals

I ∝ α

Color-coded visual · step-by-step breakdown below

  1. Torque balance: NIAB = kα
  2. I = (k/NBA)×α
  3. Calibrate scale
  4. Radial B for linearity

3. Solved Examples

Basic

Q: Double α?

Solution: I doubles

Answer: Linear

Intermediate

Q: Stronger B?

Solution: Same α needs less I

Answer: More sensitive

Advanced

Q: k increases?

Solution: Less deflection per I

Answer: Less sensitive

Exam

Q: Galvanometer relation?

Solution: I ∝ α

Answer: Sec 18.6.2

S = I_g G / (I − I_g)

Definition: Shunt resistance to convert galvanometer to ammeter.

Derivation

Parallel shunt diverts excess current; I_g through coil unchanged at full scale.

Variables

I_g = full-scale galvanometer current · G = galvanometer resistance · I = desired range

Why it works

Ammeter needs low resistance in series with circuit.

Historical context

Shunt in parallel; most current through S, small through G.

Deep understanding

Ideal ammeter: zero resistance. Effective R very small.

2. Diagrams & Visuals

S ∥ G

Color-coded visual · step-by-step breakdown below

  1. Know I_g, G, desired range I
  2. S = I_g G/(I−I_g)
  3. Connect S parallel to G
  4. Use in series in circuit

3. Solved Examples

Basic

Q: I_g=10 mA, G=100 Ω, I=1 A.

Solution: S≈1.01 Ω

Answer: ~1 Ω

Intermediate

Q: Larger range I?

Solution: S smaller

Answer: More shunt

Advanced

Q: I=I_g?

Solution: S→∞

Answer: No shunt needed

Exam

Q: Ammeter shunt?

Solution: I_g G/(I−I_g)

Answer: Sec 18.6.3

R = V/I_g − G

Definition: Series resistance to convert galvanometer to voltmeter.

Derivation

Total resistance R+G limits current to I_g at full-scale voltage V.

Variables

V = desired voltage range · R in series with G

Why it works

Voltmeter needs high resistance; connected in parallel across points.

Historical context

Most voltage drops across series R; small I_g through coil.

Deep understanding

Ideal voltmeter: infinite resistance (no circuit loading).

2. Diagrams & Visuals

R + G in series

Color-coded visual · step-by-step breakdown below

  1. Desired range V, I_g, G
  2. R + G = V/I_g
  3. R = V/I_g − G
  4. Connect across voltage points

3. Solved Examples

Basic

Q: V=10 V, I_g=1 mA, G=100 Ω.

Solution: R=9900 Ω

Answer: 9.9 kΩ

Intermediate

Q: Higher voltage range?

Solution: R larger

Answer: More series R

Advanced

Q: Voltmeter connection?

Solution: Parallel across load

Answer: Not series

Exam

Q: Voltmeter resistance?

Solution: V/I_g − G

Answer: Sec 18.6.3

5. Special Features & Extras

Complete study guide for Magnetism and Magnetic Effect of Electric Current.

Exam Tips & Tricks

  • μ₀ = 4π×10⁻⁷ T·m/A⁻¹ — always in SI magnetism problems.
  • Right-hand rule for B around wire; Fleming's left-hand for force on current/charge.
  • Long wire: B = μ₀I/(2πr) falls as 1/r.
  • Solenoid inside: B = μ₀nI uniform at centre.
  • Lorentz force does no work — speed unchanged, path curves.
  • Cyclotron ν_c independent of v — resonance condition.
  • Ammeter = shunt parallel · Voltmeter = high R series.

Common Student Mistakes

  • Using Biot-Savart when Ampere's law is simpler (symmetric cases)
  • Wrong direction with right-hand vs Fleming's rules
  • Forgetting sin θ in F = qvB and F = BIL
  • Confusing M = NIA with torque τ = MB sin θ
  • Connecting voltmeter in series instead of parallel
  • Thinking magnetic force changes kinetic energy/speed

Memory Aids & Mnemonics

Fleming's LEFT hand: Field, Current (motion), Force — motors
Parallel wires: "Same direction = friends attract" (currents attract)
Ampere's applications: Wire → 2πr · Solenoid → n · Toroid → 2πr ring
Galvanometer conversions: Ammeter = shunt ∥ · Voltmeter = R in series

Which Formula When?

  • Earth field components? → B_H = B cos δ
  • Arbitrary wire element? → Biot-Savart dB
  • Symmetric current distribution? → Ampere ∮B·dl = μ₀I
  • Long straight wire? → B = μ₀I/(2πr)
  • Solenoid/toroid? → B = μ₀nI or μ₀NI/(2πr)
  • Force on charge? → F = qvB sin θ
  • Force on wire? → F = BIL sin θ
  • Circular/helical motion? → R = mv/qB, T = 2πm/qB
  • Cyclotron? → ν_c = qB/(2πm)
  • Motor/galvanometer torque? → τ = NBIA sin θ
  • Convert galvanometer? → shunt (ammeter) or series R (voltmeter)

QUICK REFERENCE — Ch 18 Magnetism

B_H = B cos δ  |  B_V = B sin δ|dB| = (μ₀/4π)(I dl sin θ)/r²B = μ₀ N I / (2r)∮ B · dl = μ₀ IB = μ₀ I / (2πr)B = μ₀ n IB = μ₀ N I / (2πr)F = q(v × B)  |  |F| = qvB sin θF = B I L sin θF/l = μ₀ I₁ I₂ / (2πr)R = mv/(qB)  |  T = 2πm/(qB)ν_c = qB / (2πm)M = N I Aτ = M × B = N B I A sin θI = (k / N B A) × αS = I_g G / (I − I_g)R = V/I_g − G

Units: B in tesla (T) = N·A⁻¹·m⁻¹ · μ₀ = 4π×10⁻⁷ T·m/A⁻¹ · M in A·m²

Key: dB ∝ Idl sinθ/r² · B_wire = μ₀I/2πr · F = qvB · F = BIL · M = NIA

Tip: Draw field directions first with right-hand rules, then apply magnitude formulas.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L18 — Magnetism and Magnetic Effect of Electric Current only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 18 — Magnetism and Magnetic Effect of Electric Current (L18)

5 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A, Marking Scheme

Section A — Multiple Choice (1 mark)

PYQ1. A straight power line laid along east-west direction carries a current of 10 A. The earth's magnetic field at the place is 10⁻⁴ T. Force per metre experienced by the line wire will be — (A) 10⁻² N m⁻¹   (B) 10⁻³ N m⁻¹   (C) 10⁻⁴ N m⁻¹   (D) 10⁻⁵ N m⁻¹

1 mark · Section A Q15 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

Answer: (B) 10⁻³ N m⁻¹

F/L = BI sin θ. Horizontal B ⊥ east-west current → θ = 90°.

F/L = 10⁻⁴ × 10 = 10⁻³ N m⁻¹

Explanation

Force on current-carrying conductor: F = BIl sin θ. Per unit length F/L = BI when wire ⊥ B (L18 §18.4). Earth's horizontal field ⊥ vertical power line in east-west run.

Section A — Short Answer (2 marks)

PYQ2. Give any two differences between the way the electric field and the magnetic field deflect a moving charged particle.

2 marks · Section B Q35 · Marking Scheme (68/ESS/1-312-A)

Model Answer

Any two differences:

  • Electric field: FE = qE is along E — can change speed and kinetic energy.
  • Magnetic field: FB = qv × B is always ⊥ v — does no work; changes direction only (circular/helical path).
  • E can accelerate particle along field lines; B cannot increase speed, only deflects.

Explanation

Lorentz force: magnetic part always perpendicular to velocity (L18 §18.5). Electric part parallel/antiparallel to E. Cyclotron motion arises from B-only deflection.

Section B — Short Answer (2 marks)

PYQ3. A long straight wire carries a current of 3 A. Calculate the magnitude of the magnetic field at a point 10 cm away from the wire.

2 marks · Section B Q35 · 312/TUS/104A

Model Answer

B = μ₀I/(2πr)

= (4π×10⁻⁷ × 3)/(2π × 0.10)

= 6 × 10⁻⁶ T = 6 μT

Explanation

Ampere's law for long straight conductor (L18 §18.3.1): B ∝ I and B ∝ 1/r. Use r = 10 cm = 0.10 m, μ₀ = 4π×10⁻⁷ T·m·A⁻¹.

PYQ4. Find out the expression for the magnetic field due to a long solenoid carrying a current I and having n number of turns per unit length.

2 marks · Section B Q34 (OR) · Marking Scheme (68/ESS/1-312-A)

Model Answer

Consider rectangular Amperian loop through solenoid interior.

∮ B·dl = μ₀(nl)I → Bl = μ₀nIl

B = μ₀nI

Field uniform inside, parallel to axis; ≈ 0 outside.

Explanation

Marking scheme: Ampere's circuital law with enclosed current nli. Independent of solenoid radius for ideal long solenoid (L18 §18.3.2).

Section B — Long Answer (5 marks)

PYQ5. Why do two infinitely long parallel straight current-carrying conductors interact? Two 5 m long straight wires kept parallel to each other at a distance of 30 cm carry currents 10 A and 15 A in the same direction. Calculate the magnitude and direction of the force between them. Does this force tend to increase or decrease the separation between them?

5 marks · Section B Q43 · 312/MAY/204A (also Q42 · 204B/C, 68/ESS/1-312-A)

Model Answer

Why they interact: Each wire produces a magnetic field; the other wire carrying current experiences a magnetic force (Ampere's law + Lorentz force).

Force per unit length: F/L = μ₀I₁I₂/(2πd)

= (4π×10⁻⁷ × 10 × 15)/(2π × 0.30) = 10⁻⁴ N m⁻¹

Total force on 5 m length: F = 5 × 10⁻⁴ = 5 × 10⁻⁴ N

Direction: Same-direction currents → attractive (parallel wires pull together).

Separation: Force tends to decrease the distance between wires.

Explanation

Ampere's force law between parallel conductors (L18 §18.4). Opposite currents repel; same currents attract. Board numerical uses d = 30 cm = 0.30 m, L = 5 m.

Problem Solving — L18 Magnetism and Magnetic Effect of Current

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6B field

State the right-hand grip rule for direction of B around a long straight current-carrying wire.

Biot–Savart / right-hand rule

Solution — step by step with formulas

  1. Thumb along conventional current; fingers curl in direction of B lines.

Final answer: Thumb = I; fingers = B circles

Formulas used in this problem

Biot–Savart / right-hand rule

Textbook formal language

Magnetic field lines form closed loops around currents (Ampère).

Working formula set for this problem: Biot–Savart / right-hand rule. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Point thumb with the current; curled fingers show the circling field.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Magnetic field of current

B ∝ I/r for long straight wire (magnitude).

Link to chapter notes (L18 — Magnetic field of current): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Biot–Savart / right-hand rule. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write Biot–Savart / right-hand rule before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6Force on wire

Wire 0.50 m carries 2.0 A perpendicular to B = 0.40 T. Find force magnitude.

F = IlB sinθ

Solution — step by step with formulas

  1. F = 0.50×2.0×0.40 = 0.40 N.

Final answer: F = 0.40 N

Formulas used in this problem

F = IlB sinθ

Textbook formal language

Force on a straight wire is Il × B; maximum when I ⟂ B.

Working formula set for this problem: F = IlB sinθ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Current, length, and field multiply when they are at right angles.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Force on current in B

Direction by Fleming’s left-hand rule (motor rule).

Link to chapter notes (L18 — Force on current in B): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F = IlB sinθ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write F = IlB sinθ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Lorentz

Proton (q=e) moves at 10⁶ m·s⁻¹ ⟂ to B = 0.50 T. Find |F| (e=1.6×10⁻¹⁹).

F = q v B sinθ

Solution — step by step with formulas

  1. F = 1.6e-19×1e6×0.5 = 8.0×10⁻¹⁴ N.

Final answer: F = 8.0×10⁻¹⁴ N

Formulas used in this problem

F = q v B sinθ

Textbook formal language

Magnetic force on a moving charge is q(v × B); zero if v ∥ B.

Working formula set for this problem: F = q v B sinθ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Only the sideways part of velocity counts for magnetic push.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Force on charge

Magnetic force never does work (always ⟂ v).

Link to chapter notes (L18 — Force on charge): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F = q v B sinθ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write F = q v B sinθ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Circular path

Why does a charge enter a uniform B perpendicularly and move in a circle? Write r formula.

r = mv/(qB)
T = 2πm/(qB)

Solution — step by step with formulas

  1. F provides centripetal force qvB = mv²/r ⇒ r = mv/(qB).

Final answer: r = mv/(qB)

Formulas used in this problem

r = mv/(qB)
T = 2πm/(qB)

Textbook formal language

Magnetic force is perpendicular to velocity, changing direction not speed.

Working formula set for this problem: r = mv/(qB); T = 2πm/(qB). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Constant sideways shove bends the path into a circle at constant speed.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Charge in uniform B

Helix if velocity has a parallel component.

Link to chapter notes (L18 — Charge in uniform B): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: r = mv/(qB); T = 2πm/(qB). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write r = mv/(qB); T = 2πm/(qB) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Solenoid

Ideal long solenoid: n = 1000 turns/m, I = 2.0 A. Find B inside (μ₀=4π×10⁻⁷).

B = μ₀ n I (ideal, inside)

Solution — step by step with formulas

  1. B = 4πe-7 ×1000×2 ≈ 2.5×10⁻³ T.

Final answer: B ≈ 2.5 mT along axis inside

Formulas used in this problem

B = μ₀ n I (ideal, inside)

Textbook formal language

Field inside long solenoid is uniform and axial; outside ≈ 0 ideally.

Working formula set for this problem: B = μ₀ n I (ideal, inside). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Dense turns and current build a strong uniform field in the tube.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Solenoid field

Like a bar magnet’s interior field pattern.

Link to chapter notes (L18 — Solenoid field): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: B = μ₀ n I (ideal, inside). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write B = μ₀ n I (ideal, inside) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Earth B

Define magnetic declination and inclination (dip) briefly.

Solution — step by step with formulas

  1. Declination: angle between geographic and magnetic meridian.
  2. Dip: angle B makes with horizontal.

Final answer: Declination & dip describe Earth’s B orientation

Textbook formal language

Earth behaves approximately as a magnetic dipole; local field has direction parameters.

Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Compass north isn’t exactly true north (declination); field also tilts into the ground (dip).

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Earth’s magnetism

Useful in navigation and geophysics.

Link to chapter notes (L18 — Earth’s magnetism): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).