NCERT Plus Two (Class 12) Physics • Chapter 4

Moving Charges and Magnetism

Lorentz force & circular/helical paths, Biot-Savart law, Ampere's circuital law & solenoid, force between parallel wires, loop torque & moving-coil galvanometer conversion.

📄 Official NCERT Textbook PDF

Topic 4.2 – 4.3: Lorentz Force, Wire Force & Circular / Helical Paths

Why do colorful green and purple auroras dance near Earth's North and South poles while missing Kochi's equator? Because solar electrons and protons spiral along Earth's magnetic field lines in giant glowing corkscrew Helices — steered cleanly by the perpendicular Lorentz Force!
LORENTZ FORCEF = q[E + (v × B)]  |  Wire: F = I l B sinθ (1 T = 10⁴ G)
CIRCULAR (θ = 90°)r = mvqB  |  Time period T = 2πmqB (Independent of v & r!)
HELICAL (0° < θ < 90°)r = mv sinθqB  |  Pitch p = (v cosθ) T  |  Velocity filter: v = EB
🧠 Memory Trick: Magnetic force q(v × B) is ALWAYS perpendicular to velocity v ⇒ Work done W = 0 ⇒ Speed & Kinetic Energy NEVER change!
💡 Simple Kid-Check: A magnetic field acts like a polite skateboard ramp curve — it pushes strictly sideways (perpendicular to your velocity), bending your path into a circle or helix WITHOUT ever speeding you up or slowing you down!
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    When is Magnetic Force ZERO? In Fm = qvB sinθ, force vanishes if: (a) charge is stationary (v = 0), (b) neutral (q = 0), or (c) moving parallel/antiparallel to B (θ = 0°, 180°)!
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    Perpendicular Entry (θ = 90° → Circle): Magnetic force supplies centripetal force: qvB = mv²/r ⇒ r = mv/(qB). Time for one full orbit: T = 2πr/v = 2πm/(qB) (independent of v and r!).
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    Angled Entry (θ → Helix) & Velocity Selector: v⊥ = v sinθ bends into circles of radius mv sinθ/(qB) while untouched v∥ = v cosθ glides forward by pitch p = v∥ T! In crossed E ⊥ B, only speed v = E/B passes straight undeflected!
🧒 Super-Easy Comparison: A proton and an alpha particle (qα = 2e, mα = 4mp) enter the same B field perpendicularly with equal speed v. Radius ratio: rα/rp = (mα/qα)/(mp/qp) = (4/2) = 2 : 1!
Top 3 Kerala HSE traps: (1) Does a magnetic field ever change a moving charge's Kinetic Energy or speed? NO! Because Fm ⊥ v always, power Fm·v = 0 (only direction/momentum vector turns!). (2) Stationary charges feel ZERO magnetic force! (3) Time period T = 2πm/(qB) does NOT depend on speed v or radius r!

Topic 4.4 – 4.5: Biot-Savart Law & Circular Current Loop Axis Field

When Hans Christian Oersted saw a magnetic compass needle snap sideways beside a battery wire in 1820, electricity and magnetism officially became one family! Right after, Jean-Baptiste Biot and Félix Savart cracked the exact rule telling us how strong a magnetic field each tiny slice of current wire creates!
BIOT-SAVART LAWdB = μ₀4π I (dl × r̂)r²  |  μ₀4π = 10⁻⁷ T·m/A
LOOP AXIS FIELDB(x) = μ₀ N I R²2 (R² + x²)3/2 (Always along axis!)
LOOP CENTER (x = 0)B₀ = μ₀ N I2 R  |  Light speed: c = 1√(μ₀ ε₀)
🧠 Memory Trick: Check face polarity with CLOCK RULE! Look straight at the loop: Counter-Clockwise arrows form an N (North pole); Clockwise arrows form an S (South pole)!
💡 Simple Kid-Check: Unlike a point charge whose electric field shoots straight out along r̂, a tiny current element I dl creates a circular magnetic swirl perpendicular to BOTH the wire and the distance line r̂ (Curl your Right-Hand fingers around the wire!).
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    Biot-Savart Vector vs Scalar: Magnitude dB = μ₀4π I dl sinθr². Notice dB = 0 directly along the wire axis (θ = 0°, 180°) and maximum perpendicular to wire (θ = 90°)!
  • 2
    Circular Loop Axis Derivation (Top 3/5-Mark!): At distance x on axis of loop radius R, r = √(R² + x²) and dl ⊥ r (sin 90° = 1). Radial dB cosα parts from opposite elements cancel; axial components dB sinα = dB · (R/r) add up around circumference 2πR:
    B(x) = μ₀ I4π r² · (Rr) · (2πR) = μ₀ I R²2(R² + x²)3/2!
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    Special Limits: At centre (x = 0): B₀ = μ₀NI/(2R). For semicircle arc of angle α (rad): B = μ₀ I α4π R!
🧒 Super-Easy Ratio Check: At what axial distance x from a loop of radius R does field drop to 1/8 of its centre value B₀? Solve R³ / (R² + x²)3/2 = 1/8 ⇒ R / √(R² + x²) = 1/2 ⇒ R² + x² = 4R² ⇒ x = √3 R!
Watch three classic exam traps: (1) Students confuse B₀ = μ₀I/(2R) (at center of a circular loop — has NO π in denominator!) with B = μ₀I/(2πr) (near a straight long wire — HAS π in denominator!). (2) Remember c = 1/√(μ₀ε₀) = 3 × 10⁸ m/s connects electrostatics, magnetism, and light! (3) Opposite perpendicular components dB⊥ cancel on loop axis, leaving ONLY axial dB_x!

Topic 4.6 – 4.7: Ampere's Circuital Law & Solenoid Magnetic Field

Just as Gauss's Law saved pages of calculus using symmetric bags for electric fields, André-Marie Ampère found a matching lasso trick for magnetism! Loop a closed path around steady current wires, and summing B · dl around the loop directly counts the total enclosed current!
AMPERE'S LAW∮ B · dl = μ₀ Ienclosed (Valid for steady currents!)
LONG WIRE B(r)Outside (r ≥ R): B = μ₀ I2π r • Inside solid wire (r < R): B ∝ r!
LONG SOLENOIDInside: B = μ₀ n I (n = N/L turns/m!) • Ends: Bend = ½ μ₀ n I
🧠 Memory Trick: Inside a long solenoid, B = μ₀ n I — where n = N/L (turns per METRE, NOT total turns N!). Outside ideal solenoid: B = 0!
💡 Simple Kid-Check: Wind a copper wire tightly like a long spring (a Solenoid) and send current through it: all tiny loop fields team up down the hollow middle tunnel into one super-uniform magnetic highway while cancelling out outside!
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    Infinite Straight Wire (r outside): Wrap concentric circular Amperian loop of radius r: B(2πr) = μ₀ I ⇒ B = μ₀ I / (2πr)! Inside a thick solid wire of radius R, enclosed current scales as r²/R² ⇒ B(r) = μ₀Ir/(2πR²) (linear from 0 at centre!).
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    Solenoid Derivation (Top 3-Mark!): Take a rectangular Amperian loop PQRS of length L with side PQ inside along axis, RS far outside (B ≈ 0), and arms QR, SP perpendicular to B (B·dl = 0). Total enclosed current = (nL) I ⇒ B · L = μ₀ (n L) I ⇒ B = μ₀ n I!
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    End Field & Iron Core: Right at the open mouth/end of a long solenoid, only half the coils sit ahead/behind, so field drops to Bend = ½ μ₀ n I! Slide a soft-iron core inside, and field multiplies by μᵣ: B = μ₀ μᵣ n I!
🧒 Super-Easy Example: A 50 cm long solenoid has 1000 turns carrying 2 A. First convert turns per unit length: n = 1000 / 0.5 m = 2000 m⁻¹! Interior field: B = (4π × 10⁻⁷)(2000)(2) = 1.6π × 10⁻³ ≈ 5.0 × 10⁻³ T!
Watch these top 3 DHSE traps: (1) Students plug total turns N instead of turns per unit length n = N / L into B = μ₀ n I! Always divide by solenoid length in metres! (2) Notice solenoid field B = μ₀ n I is INDEPENDENT of the solenoid's cross-sectional radius (as long as length ≫ radius)! (3) At the solenoid ends, field is HALF the centre value: ½ μ₀ n I.

Topic 4.8: Force Between Parallel Currents & Definition of Ampere

Stand two extension cords side-by-side carrying heavy current: if current rushes the SAME way in both wires, they mysteriously hug together; reverse one wire's current and they push apart! That neat tug officially defines the SI unit 1 Ampere worldwide!
FORCE PER LENGTHFl = μ₀ I₁ I₂2π d (Newton's 3rd Law pair: equal & opposite!)
ATTRACT vs REPELParallel (same way) ⇒ ATTRACT! • Antiparallel (opposite) ⇒ REPEL!
1 AMPERE BENCHMARKI₁ = I₂ = 1 A at d = 1 m in vacuum ⇒ F/l = 2 × 10⁻⁷ N/m!
🧠 Memory Trick: Unlike charges, parallel currents are friendly — SAME direction ⇒ ATTRACT; OPPOSITE direction ⇒ REPEL! (Compare electron beams in free space: repulsion from static charge wins over magnetic attraction!)
💡 Simple Kid-Check: Wire 1 creates magnetic field B₁ = μ₀I₁/(2πd) right where Wire 2 sits. Then Wire 2 sitting in B₁ feels force F₂ = I₂ l B₁ — right-hand rule immediately points inward toward Wire 1 if both currents flow together!
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    Two-Step Derivation (Guaranteed 3-Mark!):
    • Wire 1 (current I₁) makes field at Wire 2 (distance d): B₁ = μ₀ I₁ / (2π d) (into page by right-hand thumb rule).
    • Force on length l of Wire 2 (F = I₂ l B₁ sin 90°): F/l = μ₀ I₁ I₂ / (2π d)!
  • 2
    Direction Rules: Same-direction currents pull toward each other (Attract); opposite-direction currents push apart (Repel). Even if I₁ = 10 A and I₂ = 1 A, both wires feel identical force magnitude (F₁₂ = F₂₁)!
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    Standard Definition of 1 Ampere (Exam Favourite!): One Ampere is that steady current which, when maintained in each of two infinitely long straight parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces between them a force equal to 2 × 10⁻⁷ N per metre of length!
🧒 Super-Easy Example: Two long parallel wires 10 cm (0.1 m) apart carry 5 A and 4 A in opposite directions. Force per metre = (2 × 10⁻⁷)(5)(4) / 0.1 = 4 × 10⁻⁵ N/m (Repulsive)!
Watch two classic traps: (1) Students mix up static charges with parallel currents: remember same-direction currents ATTRACT (whereas same-sign charges repel)! Wait: what about two parallel electron beams moving the same way in vacuum? Electrostatic repulsion (q²/4πε₀r²) is much stronger than magnetic attraction (v²/c² smaller!), so free electron beams still diverge/repel overall! (2) Always state "infinitely long, straight, negligible cross-section, 1 m apart in vacuum, 2 × 10⁻⁷ N/m" for full marks defining 1 Ampere!

Topic 4.9: Torque on Rectangular Current Loop & Magnetic Dipole Moment

Look inside a ceiling fan, mixer-grinder, or toy DC motor: at its heart sits a copper loop carrying current inside a magnet! Equal and opposite magnetic forces on its two upright arms create a nonstop twisting Torque — turning electrical current straight into spin!
MAGNETIC MOMENTm = N I A (Unit: A·m² or J/T; direction via Right-Hand Curl!)
LOOP TORQUEτ = m × B ⇒ τ = N I A B sinθ (Net force in uniform B = 0!)
ANGLE θ RULE!θ = angle between normal m & B! (If coil plane makes angle α with B, τ = NIAB cosα!)
🧠 Memory Trick: Torque hits MAXIMUM (NIAB) when the coil plane lies PARALLEL to B (because normal m is 90° to B!), and drops to ZERO when coil plane is perpendicular to B!
💡 Simple Kid-Check: A closed current loop acts like a flat bar magnet! Curl your right-hand fingers along the current around the loop — your thumb points along the North-pole normal vector m = NIA, which tries to twist and align with external field B!
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    Torque Derivation (N turns, area A = l × b): Two vertical sides of length l carry upward/downward currents perpendicular to B, feeling equal opposite forces F = N I l B separated by perpendicular lever arm b sinθ. Hence τ = (NIlB)(b sinθ) = N I A B sinθ!
  • 2
    Magnetic Dipole Moment m = N I A: Writing m = NIA simplifies torque to τ = m × B — an exact twin of electric dipole torque p × E!
  • 3
    Revolving Electron (Bohr Magneton): An electron orbiting a proton (r, speed v) forms current I = ev/(2πr), giving orbital dipole moment m = I·πr² = ½ e v r = (e2me) L. Minimum atomic quantum is Bohr magneton μB = eh/(4πme) = 9.27 × 10⁻²⁴ A·m²!
🧒 Super-Easy Angle Trap Example: A 100-turn loop (A = 0.02 m², I = 2 A) sits with its PLANE at 30° to a 0.5 T magnetic field. Then normal angle θ = 90° − 30° = 60°! Torque = NIAB sin60° = (100)(2)(0.02)(0.5)(0.866) = 1.73 N·m!
THE #1 Chapter 4 numerical trap: Read carefully whether the question gives the angle of the NORMAL to the loop (θ) OR the PLANE of the loop (α) with B! If plane is parallel to B, α = 0° ⇒ θ = 90° ⇒ τ = NIAB (MAXIMUM!). If plane is perpendicular to B, α = 90° ⇒ θ = 0° ⇒ τ = 0!

Topic 4.10: Moving Coil Galvanometer & Ammeter / Voltmeter Conversion

How does a single delicate lab galvanometer needle safely measure BOTH a 10 A car-battery surge AND a 200 V wall socket without burning up? Engineers simply team that sensitive coil up with either a tiny bypass Shunt (Ammeter) or a tall series guard Resistor (Voltmeter)!
GALVANOMETER (MCG)Radial B gives linear scale: N I A B = k φ  |  Current sens: Is = φI = NABk
→ AMMETER (Parallel S)Connect low Shunt S in PARALLEL: S = Ig GI − Ig (Ideal RA = 0!)
→ VOLTMETER (Series R)Connect high R in SERIES: R = VIg − G (Ideal RV = ∞!)
🧠 Memory Trick: A-P-L = Ammeter needs Parallel Low Shunt (S = IgG/(I−Ig)); V-S-H = Voltmeter needs Series High Resistor (R = V/Ig − G)!
💡 Simple Kid-Check: A galvanometer coil burns out if too much current enters! To turn it into a heavy-duty Ammeter, add a super-wide parallel expressway (Shunt S) so 99% of current bypasses the delicate coil. To make a Voltmeter, put a giant series roadblock (R) in front so tiny current trickles through!
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    Why Radial Magnetic Field? Concave cylindrical magnet poles + soft-iron core force field lines to stay radial, so coil plane is ALWAYS parallel to B (sin 90° = 1) at every needle angle! Thus deflecting torque NIAB balances spring restoring torque kφ ⇒ φ ∝ I (linear scale!).
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    Sensitivities & Classic Board Trap:
    • Current Sensitivity: Is = φ/I = NAB/k.
    • Voltage Sensitivity: Vs = φ/V = φ/(IRG) = NAB/(kRG).
    If you double turns N → 2N, wire length & resistance RG also double — so Is doubles, while Vs stays UNCHANGED!
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    Ammeter & Voltmeter Conversion Formulas:
    • Ammeter (range I): Same PD across G and parallel shunt S ⇒ Ig G = (I − Ig) S ⇒ S = Ig G / (I − Ig).
    • Voltmeter (range V): V = Ig(G + R) ⇒ R = (V / Ig) − G.
🧒 Super-Easy Conversion Numerical: A galvanometer (G = 20 Ω) gives full-scale deflection for Ig = 2 mA (0.002 A).
• Convert to 0–10 V voltmeter: R = (10 / 0.002) − 20 = 5000 − 20 = 4980 Ω in series!
• Convert to 0–1 A ammeter: S = (0.002 × 20) / (1 − 0.002) ≈ 0.04 Ω in parallel!
Top 3 Kerala HSE traps: (1) Does increasing turns N increase voltage sensitivity? NO! Doubling N doubles both NAB and coil resistance RG, so Vs = NAB/(kRG) stays constant! (2) Remember why a radial magnetic field & soft-iron core are used: to keep θ = 90° (τ = NIAB constant, linear scale) and strengthen B! (3) Ideal Ammeter has R = 0 (connected in series); Ideal Voltmeter has R = ∞ (connected in parallel)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Write the vector expression for Lorentz force on a moving charge. Why does a uniform magnetic field do zero work on a moving charge?

F = q[E + (v × B)].
Magnetic force Fm = q(v × B) is always perpendicular to velocity v (and displacement dr = v dt), so dW = Fm · dr = 0 ⇒ speed and KE stay constant.

Q2 (Kerala HSE Say • 3 Marks): State Biot-Savart law and apply it to derive the magnetic field along the axis of a circular current loop of radius R.

Biot-Savart: dB = (μ₀/4π) (I dl sinθ / r²).
At axial point x, distance r = √(R² + x²), dl ⊥ r. Radial components dB cosα cancel; axial components dB sinα = dB (R/r) sum over 2πR:
B = ∫ (μ₀ I dl / 4π r²)(R/r) = μ₀ I R4π r³(2πR) = μ₀ I R²2(R² + x²)3/2.

Q3 (Kerala HSE March • 3 Marks): State Ampere's circuital law. Using it, derive the magnetic field inside a long straight solenoid.

Ampere's Law: ∮ B · dl = μ₀ Ienc.
Consider rectangular Amperian loop PQRS of length L with PQ inside parallel to axis and RS outside (B ≈ 0). Side arms give 0 (B ⊥ dl).
Enclosed current = (nL)I ⇒ B L = μ₀ n L I ⇒ B = μ₀ n I.

Q4 (Kerala HSE March • 3 Marks): Derive the expression for force per unit length between two infinitely long parallel current-carrying conductors. Hence define 1 Ampere.

Wire 1 makes B₁ = μ₀I₁/(2πd) at Wire 2. Force on length l of Wire 2 is F = I₂ l B₁ = (μ₀ I₁ I₂ / 2πd) l ⇒ F/l = μ₀I₁I₂/(2πd) (attractive for same direction).
1 Ampere: Steady current in two infinitely long, negligible-section parallel wires 1 m apart in vacuum producing 2 × 10⁻⁷ N/m force.

Q5 (Kerala HSE Model • 2 Marks): Why is a radial magnetic field used in a moving coil galvanometer, and how is it produced?

Why: A radial magnetic field keeps the coil plane always parallel to B (θ = 90°, sinθ = 1) at all deflections, making torque τ = NIAB independent of φ and producing a linear scale (φ ∝ I).
How: Using concave cylindrical pole pieces and a soft-iron core inside the coil.

Q6 (Kerala HSE March • 2 Marks): Define current sensitivity of a moving coil galvanometer. Does increasing the number of turns increase voltage sensitivity? Explain.

Current sensitivity: Deflection per unit current, Is = φ/I = NAB/k.
Voltage sensitivity: Vs = φ/V = NAB/(kRG). Doubling turns N doubles coil length and resistance RG too, so N/RG stays constant — voltage sensitivity does NOT increase!

Q7 (Kerala HSE Say • 3 Marks): A galvanometer has resistance 15 Ω and gives full-scale deflection for 4 mA. How will you convert it into (a) an ammeter of range 0–6 A, and (b) a voltmeter of range 0–18 V?

(a) Ammeter (parallel shunt): S = IgG/(I − Ig) = (0.004 × 15)/(6 − 0.004) ≈ 0.01 Ω in parallel.
(b) Voltmeter (series resistance): R = V/Ig − G = (18 / 0.004) − 15 = 4500 − 15 = 4485 Ω in series.

Q8 (NCERT / DHSE Numerical • 2 Marks): An electron (m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C) moves with speed 3 × 10⁶ m/s at 60° to a magnetic field B = 0.02 T. Find its circular radius and pitch.

v⊥ = v sin60° = 2.6 × 10⁶ m/s; v∥ = v cos60° = 1.5 × 10⁶ m/s.
• Radius: r = mv⊥/(eB) ≈ 7.4 × 10⁻⁴ m (0.74 mm).
• Time period: T = 2πm/(eB) ≈ 1.79 × 10⁻⁹ s ⇒ Pitch p = v∥T ≈ 2.68 mm.