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State Board · Class 12 · Physics

Lesson 10: Magnetic Fields due to Electric Current

7 concepts · 56 practice questions with answers · Free to read

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Magnetic force on a moving charge and circular motion

From daily life

In old CRT TVs a magnet bends the electron beam.

Simple definition

The magnetic force acts only on moving charges, ⟂ to v and B : F = q v × B.

Circular motion in a magnetic field××××××××××××vFr = m v ÷ q B

In the figure : B into the page; force toward the centre ⇒ a circle.

Rule 1: Magnetic force — F = q v B sin θ (F = q v × B); force ⟂ both v and B; zero work, KE constant. 1 T = 1 N A⁻¹ m⁻¹.

  • Easy : q = 1.6 × 10⁻¹⁹ C, v = 3 × 10⁶, B = 0.2 T, θ = 90° ⇒ F = 9.6 × 10⁻¹⁴ N.
  • Medium : v ∥ B ⇒ F = 0.
  • Exam (CET) : Does the magnetic force do work? (1) no (2) yes (3) sometimes (4) maximum → answer 1)

Rule 2: Circular motion — For v ⟂ B : r = m v ÷ q B, T = 2π m ÷ q B (independent of v), ω = q B ÷ m.

  • Easy : Electron v = 2 × 10⁶, B = 0.5 mT ⇒ r = 2.28 × 10⁻² m.
  • Medium : T = 7.15 × 10⁻⁸ s.
  • Exam (CET) : T is independent of? (1) v (2) m (3) q (4) B → answer 1)

Rule 3: Energy and momentum — r = p ÷ qB; KE = q² B² r² ÷ 2m. For equal KE, r ∝ √m ÷ q.

  • Easy : Proton and α at equal speed : r_α : r_p = 2 : 1 (m 4×, q 2×).
  • Medium : Equal KE : r_α = r_p.
  • Exam (CET) : Proton and deuteron (same v) : r_d : r_p? (1) 2 (2) 1 (3) 4 (4) 1/2 → answer 1)

Derivation : r = m v ÷ q B

  • Step 1 : F = q v B provides the centripetal force.
  • Step 2 : m v² ÷ r = q v B.
  • Step 3 : Solving : r = m v ÷ q B.
  • Step 4 : T = 2π r ÷ v = 2π m ÷ q B.
Remember

F = qvB sinθ; r = mv/qB; T = 2πm/qB.

Common mistake / CET trap

The force does no work; speed is unchanged, direction changes.

Shortcut

Memory clue : "perpendicular force = circle".

Spot the difference

QuantityFormulaUnit / dimensions
FqvB sinθN
rmv/qBm
T2πm/qBs
BF/qvT; [M T⁻² A⁻¹]

Practice questions

Easy
  1. 1Unit of B?
  2. 2Direction of force?
  3. 3T does not depend on?
Medium
  1. 4q = 2 µC, v = 10³, B = 0.5 T : F?
  2. 5Electron r = 2 cm; B doubled : r?
  3. 6m = 2 × 10⁻²⁷, q = 1.6 × 10⁻¹⁹, B = 0.1 T : T?
Challenging
  1. 7α and proton, equal KE : ratio of radii?
  2. 8v at 30° to B : F relative to 90°?
Show answers (try them yourself first)
  1. 1Tesla
  2. 2⟂ to v and B
  3. 3v
  4. 4F = 1 × 10⁻³ N
  5. 5r = 1 cm
  6. 6T = 7.85 × 10⁻⁷ s
  7. 71 : 1
  8. 81/2

Cyclotron, helical motion and crossed fields

From daily life

Pushing a swing at the right moment builds the motion; a cyclotron does it to particles.

Simple definition

A cyclotron accelerates particles using a magnetic field and an oscillating electric field.

Cyclotron : the two D-shaped deesf = qB ÷ 2π mD₁D₂

In the figure : Half circles in D₁ and D₂; speed gained at the gap.

Live figure

Cyclotron, helical motion and crossed fields

In the app you can move this figure with your finger.

Open in the app

Rule 1: Cyclotron principle — T is independent of v and r ⇒ fixed frequency f = q B ÷ 2π m; the electric field in the gap boosts the particle each half turn.

  • Easy : Proton, B = 1.5 T ⇒ f = 22.9 MHz.
  • Medium : Match the oscillator frequency to f.
  • Exam (CET) : Cyclotron frequency depends on? (1) q, B, m (2) v (3) r (4) E → answer 1)

Rule 2: Maximum energy — KE_max = q² B² R² ÷ 2m.

  • Easy : B = 1.5 T, R = 0.5 m, proton ⇒ KE = 26.9 MeV.
  • Medium : R doubled ⇒ KE four times.
  • Exam (CET) : KE_max ∝ ? (1) R² (2) R (3) 1/R (4) R⁰ → answer 1)

Rule 3: Helical motion and crossed fields — A component of v along B gives a helix; pitch = v_∥ T. With E ⟂ B, the velocity selector passes v = E ÷ B straight.

  • Easy : E = 2 × 10⁴ V/m, B = 0.04 T ⇒ v = 5 × 10⁵ m/s.
  • Medium : Pitch = 2π m v cosθ ÷ qB.
  • Exam (CET) : Selector speed v? (1) E/B (2) B/E (3) EB (4) E−B → answer 1)

Derivation : KE_max = q²B²R² ÷ 2m

  • Step 1 : At outer radius R : v = q B R ÷ m.
  • Step 2 : KE = ½ m v².
  • Step 3 : = ½ m (qBR/m)².
  • Step 4 : KE = q² B² R² ÷ 2m.
Remember

f = qB/2πm; KE_max = q²B²R²/2m; v = E/B.

Common mistake / CET trap

KE_max depends on mass, not on E.

Shortcut

Memory clue : "frequency fixed, energy ∝ R²".

Spot the difference

DeviceFormulaCondition
Cyclotron fqB/2πmconstant
KE_maxq²B²R²/2mR = dee radius
Velocity selectorE/BE ⟂ B ⟂ v

Practice questions

Easy
  1. 1What does a cyclotron accelerate?
  2. 2f depends on?
  3. 3Selector formula?
Medium
  1. 4B = 1 T, α : f? (m = 6.64 × 10⁻²⁷, q = 3.2 × 10⁻¹⁹)
  2. 5E = 10⁴, B = 0.02 : v?
  3. 6R = 0.4, B = 1 T proton : KE?
Challenging
  1. 7Pitch : v = 10⁶, θ = 60°, B = 0.01 T, electron
  2. 8KE if R doubled?
Show answers (try them yourself first)
  1. 1Charged particles
  2. 2q, B, m
  3. 3E/B
  4. 4f = 7.67 MHz
  5. 5v = 5 × 10⁵ m/s
  6. 6KE = 7.66 MeV
  7. 7Pitch = 1.79 × 10⁻³ m
  8. 84 times

Force on current-carrying conductors and between parallel wires

From daily life

An electric motor turns because of the force on wires in a magnetic field.

Simple definition

The force on a wire is the sum of forces on many moving charges.

Force between parallel wiresFF/l = μ₀I₁I₂ ÷ 2π dI₁I₂

In the figure : Parallel currents ⇒ attraction.

Rule 1: Force on a current-carrying wire — F = I l B sin θ (F = I l × B). Direction : Fleming's left-hand rule.

  • Easy : I = 5 A, l = 0.2 m, B = 0.4 T, θ = 30° ⇒ F = 0.2 N.
  • Medium : I ∥ B ⇒ F = 0.
  • Exam (CET) : F is maximum when? (1) θ = 90° (2) 0° (3) 45° (4) 180° → answer 1)

Rule 2: Parallel wires — Currents in the same direction attract; opposite repel. F ÷ l = μ₀ I₁ I₂ ÷ 2π d.

  • Easy : I₁ = I₂ = 5 A, d = 0.1 m ⇒ F/l = 5 × 10⁻⁵ N/m.
  • Medium : d doubled ⇒ F halves.
  • Exam (CET) : Parallel currents wires? (1) attract (2) repel (3) zero (4) rotate → answer 1)

Rule 3: Definition of the ampere — In vacuum, two long wires 1 m apart experiencing 2 × 10⁻⁷ N per metre carry 1 A each.

  • Easy : 1 A, 1 m : F/l = 2 × 10⁻⁷ N/m ✓.
  • Medium : μ₀ = 4π × 10⁻⁷ T m A⁻¹.
  • Exam (CET) : F/l = 2 × 10⁻⁷ N/m ⇒ I = ? (1) 1 A (2) 2 A (3) 0.5 A (4) 10 A → answer 1)

Derivation : F/l = μ₀ I₁ I₂ ÷ 2π d

  • Step 1 : Field of wire 1 at wire 2 : B = μ₀ I₁ ÷ 2π d.
  • Step 2 : Force on wire 2 : F = I₂ l B.
  • Step 3 : F = I₂ l μ₀ I₁ ÷ 2π d.
  • Step 4 : F ÷ l = μ₀ I₁ I₂ ÷ 2π d.
Remember

F = IlBsinθ; F/l = μ₀I₁I₂/2πd.

Common mistake / CET trap

Parallel currents attract (opposite to like charges!).

Shortcut

Memory clue : "same way = together".

Spot the difference

QuantityFormulaUnit
FIlB sinθN
F/lμ₀I₁I₂/2πdN/m
μ₀4π×10⁻⁷T m A⁻¹

Practice questions

Easy
  1. 1Formula for F?
  2. 2Same-direction currents?
  3. 3Define ampere.
Medium
  1. 4I = 8, l = 0.5, B = 0.2, θ = 90° : F?
  2. 5I₁ = 10, I₂ = 20, d = 0.2 : F/l?
  3. 6d tripled : F/l?
Challenging
  1. 7Wires 4 cm apart, F/l = 5 × 10⁻⁵, I₁ = I₂ : I?
  2. 8Both currents doubled : F/l?
Show answers (try them yourself first)
  1. 1I l B sin θ
  2. 2Attraction
  3. 32 × 10⁻⁷ N/m ⇒ 1 A
  4. 4F = 0.8 N
  5. 5F/l = 2 × 10⁻⁴ N/m
  6. 61/3
  7. 7I = 3.2 A
  8. 84 times

Torque, magnetic dipole and the moving coil galvanometer

From daily life

A motor's armature turns due to the torque on a loop.

Simple definition

The magnetic dipole moment is m = N I A.

Moving coil galvanometerNSτ = NIAB sinθθ ÷ I = NAB ÷ k

In the figure : In a radial field the deflection ∝ current.

Rule 1: Torque on a loop — τ = N I A B sin θ (θ = angle between area vector and B); magnetic moment m = N I A, τ = m B sin θ.

  • Easy : N = 100, I = 10 mA, A = 4 cm², B = 0.2 T ⇒ τ(max) = 8 × 10⁻⁵ N m.
  • Medium : m = N I A = 100 × 0.01 × 4 × 10⁻⁴ = 4 × 10⁻⁴ A m².
  • Exam (CET) : τ is maximum when? (1) B in the plane (2) B ⟂ plane (3) 45° (4) never → answer 1)

Rule 2: Moving coil galvanometer — A radial field makes τ = N I A B always; restoring torque k θ ⇒ θ = N A B I ÷ k (deflection ∝ I).

  • Easy : N = 50, A = 3 × 10⁻⁴, B = 0.1, k = 10⁻⁷ : θ/I = 50 × 3 × 10⁻⁴ × 0.1 ÷ 10⁻⁷ = 1.5 × 10⁴ rad/A.
  • Medium : A soft iron core makes the field radial.
  • Exam (CET) : Deflection ∝ ? (1) I (2) I² (3) √I (4) 1/I → answer 1)

Rule 3: Magnetic dipole — m = N I A; in a field U = −m B cos θ; τ = m × B. A bar magnet and a loop behave alike.

  • Easy : m = 0.5 A m², B = 0.2 T, θ = 60° : τ = 0.0866 N m.
  • Medium : U(0°) = −m B.
  • Exam (CET) : Unit of m? (1) A m² (2) A/m (3) N m (4) T → answer 1)

Derivation : τ = N I A B sin θ

  • Step 1 : Rectangular loop a × b; force on the sides F = I a B.
  • Step 2 : Two equal opposing forces form a couple; arm b sin θ.
  • Step 3 : τ = I a B b sin θ = I A B sin θ.
  • Step 4 : N turns : τ = N I A B sin θ.
Remember

τ = NIAB sinθ; m = NIA; θ = NABI/k.

Common mistake / CET trap

θ is with the area vector, not the plane; with the plane use cos.

Shortcut

Memory clue : "τ = m × B" (m = NIA).

Spot the difference

QuantityFormulaUnit
τNIAB sinθN m
mNIAA m²
θ/INAB/krad/A

Practice questions

Easy
  1. 1Formula for τ?
  2. 2Unit of m?
  3. 3Why radial field?
Medium
  1. 4N = 20, I = 0.5, A = 0.01, B = 0.3, θ = 90° : τ?
  2. 5m = 2 A m², B = 0.5 T, θ = 30° : τ?
  3. 6Loop radius 10 cm, I = 2 A, N = 1 : m?
Challenging
  1. 7I needed to double θ?
  2. 8Change in energy θ = 0 → 180°?
Show answers (try them yourself first)
  1. 1NIAB sinθ
  2. 2A m²
  3. 3To keep τ constant
  4. 4τ = 0.03 N m
  5. 5τ = 0.5 N m
  6. 6m = 0.0628 A m²
  7. 7Double I
  8. 82 m B

Biot–Savart law, long wire and circular loop

From daily life

A compass near a wire deflects — a current produces a magnetic field.

Simple definition

The Biot–Savart law gives the field of a current element.

Biot–Savart : wire and circular loopIB = μ₀I ÷ 2πrB = μ₀I ÷ 2RR

In the figure : Circles around the wire; at a loop's centre B = μ₀I/2R.

Rule 1: Biot–Savart law — dB = (μ₀ ÷ 4π) · I dl sin θ ÷ r². Direction : right-hand rule.

  • Easy : I dl = 10⁻² A m, r = 0.1 m, θ = 90° ⇒ dB = 10⁻⁷ × 10⁻² ÷ 10⁻² = 10⁻⁷ T.
  • Medium : θ = 0 (along the element) ⇒ dB = 0.
  • Exam (CET) : μ₀/4π = ? (1) 10⁻⁷ (2) 10⁻⁶ (3) 4π × 10⁻⁷ (4) 1 → answer 1)

Rule 2: Long straight wire — B = μ₀ I ÷ 2π r; field lines are circles. r doubled ⇒ B halves.

  • Easy : I = 10 A, r = 5 cm ⇒ B = 4 × 10⁻⁵ T.
  • Medium : Two wires with opposite currents : fields add at the midpoint.
  • Exam (CET) : B ∝ ? (1) 1/r (2) 1/r² (3) r (4) r² → answer 1)

Rule 3: Circular loop — At the centre B = μ₀ N I ÷ 2R; on the axis B = μ₀ N I R² ÷ 2(R² + x²)^{3/2}.

  • Easy : I = 2 A, R = 10 cm ⇒ B = 1.26 × 10⁻⁵ T.
  • Medium : N = 50 ⇒ B = 6.28 × 10⁻⁴ T.
  • Exam (CET) : On the axis at x = 0 ⇒ B? (1) μ₀NI/2R (2) 0 (3) μ₀I/2πR (4) infinite → answer 1)

Derivation : B at the centre of a loop

  • Step 1 : Each dl ⟂ r, r = R : dB = (μ₀/4π) I dl ÷ R².
  • Step 2 : All dB point the same way ⇒ add.
  • Step 3 : ∮ dl = 2πR.
  • Step 4 : B = (μ₀/4π) I 2πR ÷ R² = μ₀ I ÷ 2R.
Remember

B = μ₀I/2πr (wire); μ₀NI/2R (loop).

Common mistake / CET trap

2πr for a wire, 2R for a loop — do not mix.

Shortcut

Memory clue : "wire 2πr, loop 2R".

Spot the difference

ShapeBUnit
Long wireμ₀I/2πrT
Loop centreμ₀NI/2RT
Semicircle centreμ₀I/4RT

Practice questions

Easy
  1. 1State Biot–Savart law.
  2. 2Shape of field lines of a wire?
  3. 3Value of μ₀?
Medium
  1. 4I = 5 A, r = 0.1 m : B?
  2. 5Loop R = 5 cm, I = 1 A, N = 20 : B?
  3. 6r tripled : B?
Challenging
  1. 7Semicircle R = 10 cm, I = 4 A : B?
  2. 8Two wires 10 cm apart, opposite 5 A, B at the midpoint?
Show answers (try them yourself first)
  1. 1dB = (μ₀/4π) Idl sinθ/r²
  2. 2Yes
  3. 34π × 10⁻⁷ T m A⁻¹
  4. 4B = 1 × 10⁻⁵ T
  5. 5B = 2.51 × 10⁻⁴ T
  6. 61/3
  7. 7B = 1.26 × 10⁻⁵ T
  8. 8B = 4 × 10⁻⁵ T

Ampere's circuital law

From daily life

Gauss's law for E; Ampere's law for B — equally convenient.

Simple definition

Ampere's law links ∮ B · dl around a closed loop to the enclosed current.

Ampere's circuital lawIB∮ B · dl = μ₀ I

In the figure : Circular loop; B tangential.

Rule 1: Ampere's law — ∮ B · dl = μ₀ I_enclosed; for symmetric distributions B is found easily.

  • Easy : I = 15 A enclosed ⇒ ∮ B · dl = 4π × 10⁻⁷ × 15 = 1.88 × 10⁻⁵ T m.
  • Medium : Outside currents are not counted.
  • Exam (CET) : ∮ B·dl = ? (1) μ₀I (2) I/μ₀ (3) μ₀/I (4) I → answer 1)

Rule 2: Field of a thick wire — Radius a : outside B = μ₀I/2πr; inside B = μ₀ I r ÷ 2π a² (r < a). Maximum at the surface.

  • Easy : I = 20 A, a = 2 cm, r = 4 cm ⇒ B = 1 × 10⁻⁴ T.
  • Medium : B = 0 at the centre.
  • Exam (CET) : Inside a thick wire B ∝ ? (1) r (2) 1/r (3) r² (4) constant → answer 1)

Rule 3: Choosing an amperian loop — Pick a loop matching the symmetry : B tangential and constant on it; ∮ B · dl = B L.

  • Easy : A circle for a wire; a rectangle for a solenoid.
  • Medium : Where B ⟂ dl the contribution is 0.
  • Exam (CET) : Best loop for a wire? (1) circle (2) square (3) triangle (4) line → answer 1)

Derivation : B of a long wire (Ampere)

  • Step 1 : Circular loop of radius r; B tangential and constant.
  • Step 2 : ∮ B · dl = B (2πr).
  • Step 3 : = μ₀ I.
  • Step 4 : B = μ₀ I ÷ 2π r.
Remember

∮B·dl = μ₀I; B = μ₀I/2πr.

Common mistake / CET trap

Ignore currents outside; sign of enclosed current by the right-hand rule.

Shortcut

Memory clue : "Gauss = E, Ampere = B".

Spot the difference

PositionBCondition
Outsideμ₀I/2πrr > a
Insideμ₀Ir/2πa²r < a
Centre0r = 0

Practice questions

Easy
  1. 1State Ampere's law.
  2. 2B at the wire's axis?
  3. 3Choice of loop?
Medium
  1. 4I = 30 A, r = 6 cm (outside) : B?
  2. 5a = 1 cm, I = 10 A, r = 0.5 cm : B?
  3. 6B at the surface vs at r = 2a?
Challenging
  1. 7At what r is B = 10⁻⁴ T for I = 100 A?
  2. 8Why B ∝ r inside?
Show answers (try them yourself first)
  1. 1∮B·dl = μ₀I
  2. 2Zero
  3. 3One matching the symmetry
  4. 4B = 1 × 10⁻⁴ T
  5. 5B = 1 × 10⁻⁴ T
  6. 62 : 1
  7. 7r = 20.0 cm
  8. 8Enclosed current ∝ r²

Solenoid and toroid

From daily life

A doorbell, relay or crane magnet all use a solenoid.

Simple definition

A solenoid is wire wound on a cylinder; a toroid on a ring.

Solenoid and toroidB = μ₀ n IB = μ₀NI ÷ 2πr

In the figure : Uniform B inside; in a toroid B is circular.

Rule 1: Solenoid — Inside a long solenoid B = μ₀ n I (n = turns/length), ≈ 0 outside; half at the ends.

  • Easy : n = 500 turns/m, I = 2 A ⇒ B = 1.26 × 10⁻³ T.
  • Medium : At the end B = μ₀nI/2.
  • Exam (CET) : B inside a solenoid? (1) uniform (2) varying (3) zero (4) ∝ r → answer 1)

Rule 2: Toroid — N turns, mean radius r : B = μ₀ N I ÷ 2π r; zero in the hole and outside.

  • Easy : N = 400, I = 3 A, r = 10 cm ⇒ B = 2.4 × 10⁻³ T.
  • Medium : n = N ÷ 2πr ⇒ B = μ₀nI.
  • Exam (CET) : B outside a toroid? (1) 0 (2) maximum (3) μ₀nI (4) infinite → answer 1)

Rule 3: Iron core — A soft iron core multiplies B : B = μ₀ μ_r n I (μ_r = relative permeability).

  • Easy : μ_r = 200 ⇒ B becomes 200 times.
  • Medium : Electromagnets : cranes, bells.
  • Exam (CET) : With an iron core B? (1) increases (2) decreases (3) same (4) zero → answer 1)

Derivation : Solenoid B = μ₀ n I

  • Step 1 : Rectangular amperian loop with length l inside.
  • Step 2 : ∮ B · dl = B l (other sides give 0).
  • Step 3 : I_enclosed = n l I.
  • Step 4 : B l = μ₀ n l I ⇒ B = μ₀ n I.
Remember

B = μ₀nI (solenoid); μ₀NI/2πr (toroid).

Common mistake / CET trap

n = N/l (turns per metre); do not use N alone.

Shortcut

Memory clue : "inside uniform, outside zero".

Spot the difference

DeviceB (inside)B (outside)
Solenoidμ₀nI≈ 0
Toroidμ₀NI/2πr0
Wireμ₀I/2πr—

Practice questions

Easy
  1. 1B of a solenoid?
  2. 2What is n?
  3. 3B outside a toroid?
Medium
  1. 4n = 1000/m, I = 0.5 A : B?
  2. 5N = 200, r = 5 cm, I = 1 A : B?
  3. 6n doubled, I halved : B?
Challenging
  1. 7B = 0.02 T, I = 4 A : n?
  2. 8μ_r = 500, n = 200, I = 1 A : B?
Show answers (try them yourself first)
  1. 1μ₀nI
  2. 2Turns per length
  3. 3Zero
  4. 4B = 6.28 × 10⁻⁴ T
  5. 5B = 8 × 10⁻⁴ T
  6. 6Unchanged
  7. 7n = 3979 /m
  8. 8B = 1.26 × 10⁻¹ T

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
ForceF = qvB sinθ; F = IlB sinθN
Circler = mv/qB; T = 2πm/qBm; s
Cyclotronf = qB/2πmHz
WiresF/l = μ₀I₁I₂/2πdN/m
Loopτ = NIAB sinθ; m = NIAN m; A m²
WireB = μ₀I/2πrT
Loop Bμ₀NI/2RT
SolenoidB = μ₀nIT; [M T⁻² A⁻¹]

🎯 Most important points for the exam

  • F ⟂ v ⇒ zero work; r = mv/qB; T independent of v.
  • Cyclotron f = qB/2πm; KE_max = q²B²R²/2m.
  • Parallel currents attract; F/l = μ₀I₁I₂/2πd.
  • τ = NIAB sinθ; galvanometer θ ∝ I.
  • Biot–Savart : wire 2πr, loop 2R.
  • Ampere : ∮B·dl = μ₀I_enclosed.
  • Solenoid μ₀nI; toroid μ₀NI/2πr.
Study mantra

Vector direction (right hand) → correct formula → SI units → answer with direction.