Moving Charges and Magnetism - Class 12 Physics

Class 12 Physics | Unit I — Magnetism

Chapter 4: Moving Charges and Magnetism

Biot-Savart Law • Ampere's Law • Lorentz Force • Solenoid • Toroid • Cyclotron • Galvanometer

1. Magnetic Force on a Moving Charge (Lorentz Force)

A magnetic field B exerts a force on a charged particle only when it is moving. This force is called the magnetic Lorentz force. Unlike electric force, magnetic force does no work.

F = q(v × B) | Magnitude: F = qvB sinθ
Total Lorentz Force: F = q(E + v × B)
q = charge (C), v = velocity (m/s), B = magnetic field (T), θ = angle between v and B

1.1 Key Properties of Magnetic Force

Direction: perpendicular to both v and B (right-hand rule / cross product).
F = 0 when θ = 0° or 180° (v parallel or anti-parallel to B).
F is maximum when θ = 90° (v ⊥ B): F_max = qvB.
Work done by magnetic force = 0 always (F ⊥ displacement). Therefore, magnetic force cannot change the kinetic energy or speed of the particle — only direction.
Magnetic force does not change the magnitude of velocity, only its direction → particle moves in a circle when v ⊥ B.
Direction for positive charge: Fleming's Left-Hand Rule (thumb = force F, index = B, middle = I or v). For negative charges, reverse the force direction.

1.2 Circular Motion of Charged Particle in Uniform B

When a charged particle enters a uniform magnetic field perpendicular to B, the magnetic force acts as centripetal force, causing uniform circular motion.

Centripetal force = Magnetic force:
mv²/r = qvB ⇒ r = mv/(qB) = p/(qB)

Time period: T = 2πr/v = 2πm/(qB) — independent of v and r!
Frequency: f = qB/(2πm) — cyclotron frequency, independent of speed.
Angular velocity: ω = qB/m

r ∝ momentum (p = mv): heavier or faster particle has larger radius.
T is constant for given q, m, B — this is the principle of the cyclotron.
For helical motion (v has components both parallel and perpendicular to B): particle traces a helix. Pitch = v_∥ × T.

⚠️ NEET (2015, 2018, 2020, 2022): Work done by B on charge = 0 always. r = mv/qB. If same kinetic energy: r ∝ √m/q. If same momentum: r ∝ 1/q. T = 2πm/qB (no v dependence). Proton and alpha particle: same kinetic energy → r_α/r_p = √(m_α/q_α) / √(m_p/q_p) = √(4m/2e ÷ m/e) = √2.

2. Biot-Savart Law

Biot-Savart Law gives the magnetic field dB produced by a small current element I dl at a point P at distance r. It is the magnetic analogue of Coulomb's law for electric field.

dB = (μ₀/4π) × (I |dl × r̂|) / r² = (μ₀/4π) × (I dl sinθ) / r²
μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space; μ₀/4π = 10⁻⁷ T·m/A)

dB is perpendicular to both dl and r (right-hand rule).
dB ∝ I, dB ∝ dl, dB ∝ sinθ, dB ∝ 1/r².
dB = 0 along the line of current (θ = 0 or 180°).

2.1 B at Centre of a Circular Loop

Single circular loop (radius R, current I):
B = μ₀I / (2R)

For N turns: B = Nμ₀I / (2R)
Direction: along the axis, given by right-hand curl rule (curl fingers in direction of current → thumb points in direction of B).

2.2 B at Axial Point of a Circular Loop

At distance x on axis from centre:
B = μ₀IR² / [2(R² + x²)^3/2]
At x = 0 (centre): B = μ₀I/2R ✓
At x ≫ R (far away): B ≈ μ₀IR²/(2x³) (like a magnetic dipole)

2.3 B Due to a Straight Current-Carrying Wire

Finite wire, at perpendicular distance d:
B = (μ₀I / 4πd)(sinθ₁ + sinθ₂)

Infinite straight wire: B = μ₀I / (2πd)
Semi-infinite wire: B = μ₀I / (4πd)

⚠️ NEET: B at centre of loop = μ₀NI/(2R). B of infinite wire = μ₀I/(2πd). B on axis at x: denominator (R²+x²)^3/2. Current in same direction in two coaxial loops → fields add. Opposite → fields subtract.

3. Ampere's Circuital Law and Applications

Ampere's Circuital Law: The line integral of magnetic field B along a closed loop (Amperian loop) equals μ₀ times the total current passing through the surface bounded by that loop.

∮ B · dl = μ₀ I_enc
(Closed line integral of B around Amperian loop = μ₀ × net enclosed current)

3.1 B Due to a Long Straight Wire (from Ampere's Law)

Choose circular Amperian loop of radius r around wire:
B(2πr) = μ₀I ⇒ B = μ₀I / (2πr) (same result as Biot-Savart)

3.2 Inside a Solenoid

B = μ₀nI (uniform, inside)
n = N/L = number of turns per unit length
B = 0 outside an ideal infinite solenoid
For finite solenoid: B at end = μ₀nI/2

Solenoid acts like a bar magnet: south face → current flows clockwise; north face → anticlockwise (when viewed from outside).
Field inside is uniform and very strong. Used in electromagnets, MRI machines.
Field outside solenoid ≈ 0 (fields of adjacent turns cancel).

3.3 Inside a Toroid

Inside toroid (mean radius r₀, N turns):
B = μ₀NI / (2πr₀)
Outside and inside the hole of toroid: B = 0

Toroid is a solenoid bent into a ring. Magnetic field is completely confined within the toroid — no leakage.
Used in transformers, inductors in electronics.

⚠️ NEET: Solenoid B = μ₀nI (n = turns/length). At one end of solenoid = μ₀nI/2. Toroid: B outside = 0, B inside = μ₀NI/2πr. Both use Ampere's law. Difference: solenoid is open-ended; toroid is closed.

4. Force on a Current-Carrying Conductor in Magnetic Field

A current-carrying conductor placed in a magnetic field experiences a force. This is because the moving charges (current) experience Lorentz force = F on each charge → net force on conductor.

F = I(L × B) | Magnitude: F = BIL sinθ
F is maximum when I ⊥ B (θ=90°): F = BIL
F = 0 when I ∥ B (θ=0°)

Direction: Fleming's Left-Hand Rule (for conventional current).
For arbitrarily shaped conductor in uniform B: F = I(L_eff × B) where L_eff = straight-line distance between endpoints.
Closed loop in uniform B: net force = 0 (but torque may exist).

4.1 Force Between Two Parallel Current-Carrying Wires

Wire 1 creates field B₁ = μ₀I₁/(2πd) at wire 2.
Force per unit length on wire 2: f = μ₀I₁I₂ / (2πd)

Same direction currents: attract
Opposite direction currents: repel

Definition of 1 Ampere: 1 A is the current that, flowing through each of two long parallel wires 1 m apart in vacuum, produces a force of exactly 2 × 10⁻⁷ N per metre between them.

5. Torque on a Current Loop in Magnetic Field

Torque on a rectangular loop (N turns, area A) in uniform B:
τ = NIAB sinθ
In vector form: τ = m × B where m = NIA (magnetic dipole moment)
θ = angle between magnetic moment m and B

Magnetic dipole moment: m = NIA. Direction: normal to plane of coil (right-hand rule). Unit: A·m².
Torque is maximum (θ = 90°, plane of loop parallel to B): τ_max = NIAB.
Torque is zero (θ = 0° or 180°, plane ⊥ B, m ∥ or anti-∥ B): equilibrium position.
θ = 0°: stable equilibrium (m parallel to B, energy minimum = −mB).
θ = 180°: unstable equilibrium (m anti-parallel to B, energy maximum = +mB).
Potential energy: U = −m·B = −mB cosθ.

6. Moving Coil Galvanometer (MCG)

Moving Coil Galvanometer: An instrument that detects and measures small electric currents. It uses the torque on a current-carrying coil in a magnetic field.

6.1 Working Principle

Coil suspended in a radial magnetic field (concave cylindrical pole pieces + soft iron core). In radial field, plane of coil is always parallel to B → θ = 90° always.
Deflecting torque = τ_d = NIAB (constant for given I).
Restoring torque from phosphor-bronze hair spring: τ_r = kθ (k = spring constant, θ = deflection).
At equilibrium: NIAB = kθ ⇒ θ = (NBA/k)I
θ ∝ I → uniform (linear) scale. This is the key advantage of radial field.

Current sensitivity: θ/I = NBA/k (rad/A or degree/A)
Voltage sensitivity: θ/V = NBA/(kR_g)
Increase sensitivity: increase N, B, A; decrease k (softer spring).

6.2 Conversion to Ammeter

Connect low resistance shunt S in parallel with galvanometer.
I_gR_g = (I − I_g)S ⇒ S = I_gR_g / (I − I_g)
Effective resistance: R_A = R_gS/(R_g+S) (very small)

Ammeter has very low resistance → connected in series in circuit. Ideal ammeter: R = 0.
Shunt S is made of low resistance material (manganin) for stability with temperature.

6.3 Conversion to Voltmeter

Connect high resistance R in series with galvanometer.
V = I_g(R_g + R) ⇒ R = V/I_g − R_g
Effective resistance: R_V = R_g + R (very large)

Voltmeter has very high resistance → connected in parallel across component. Ideal voltmeter: R = ∞.

⚠️ NEET (2014, 2017, 2019, 2021): Galvanometer: radial field → θ ∝ I (linear scale). θ = NIAB/k. Sensitivity increased by: ↑ N, ↑ B, ↑ A, ↓ k. Ammeter: shunt in parallel (S is small). Voltmeter: R in series (R is large). Ideal ammeter R = 0; Ideal voltmeter R = ∞. Changing R in voltmeter does NOT change sensitivity (changes range only).

7. Cyclotron

Cyclotron: A device to accelerate charged particles (like protons, deuterons, alpha particles) to very high kinetic energies using repeated action of electric field and circular motion in a magnetic field. Invented by Ernest O. Lawrence (1930).

7.1 Construction and Working

Two D-shaped hollow metallic chambers called dees (D₁ and D₂), separated by a narrow gap. High-frequency alternating voltage applied across the gap.
Strong uniform magnetic field B perpendicular to the plane of dees (provided by powerful electromagnet).
Charged particle placed at centre. Electric field (in gap) accelerates it. Inside dee, no E field → particle moves in semicircle due to B.
Each time particle crosses gap, electric field reverses polarity (frequency = cyclotron frequency) → particle gains energy each crossing.
Radius of semicircular path increases with each cycle. Particle spirals outward until it reaches rim of dee → deflected out by deflector plate.

Cyclotron (resonance) condition: oscillator frequency = cyclotron frequency
ν_c = qB / (2πm) (independent of v and r — isochronous)

Maximum kinetic energy:
KE_max = q²B²R² / (2m) (R = radius of dee)

Maximum velocity: v_max = qBR/m

7.2 Limitations of Cyclotron

Cannot accelerate neutral particles (neutrons, photons).
Cannot accelerate electrons effectively: they become relativistic even at low energies. Relativistic mass increase changes ν_c, breaking resonance condition.
At very high energies, relativistic mass increase causes resonance to fail. Solution: synchrocyclotron (modulated frequency) or synchrotron.
Maximum energy limited by radius of dees.

⚠️ NEET (2013, 2016, 2019, 2022): Cyclotron frequency ν_c = qB/2πm (no v, no r — isochronous!). KE_max = q²B²R²/2m. Cannot accelerate: neutrons, electrons (relativistic). Time period T = 2πm/qB (constant). Lawrence invented cyclotron (1930). Dees are hollow (no electric field inside).

8. Important Comparative Facts

Property	Electric Force (qE)	Magnetic Force (qv×B)
Acts on	All charges (rest or moving)	Only moving charges
Direction	Along E (or opposite for −q)	Perpendicular to v and B
Work done	Can do work (changes KE)	Zero (only changes direction)
Effect on speed	Changes speed	Does NOT change speed
Path of particle	Parabolic (if ⊥ to initial v)	Circular or helical
Depends on	Charge q only	Charge, velocity, angle

Comparison: Biot-Savart vs Ampere's Law

Feature	Biot-Savart Law	Ampere's Circuital Law
Use	Any current distribution	Symmetrical current distributions only
Calculation	Integration over current element	Line integral along closed loop
Preferred for	Circular loops, curved wires	Long straight wire, solenoid, toroid
Analogy	Coulomb's law (field due to source)	Gauss's law (flux through surface)

🎓 NEET Previous Year Questions

Q1. [NEET 2022] A proton and an alpha particle move with the same kinetic energy in a magnetic field. The ratio of radii of their circular paths (r_p : r_α) is:

Answer KE = p²/(2m) ⇒ p = √(2mKE). r = p/(qB) = √(2mKE) / (qB)
r_p/r_α = [√(2m_pKE) / q_pB] / [√(2m_αKE) / q_αB]
= (q_α/q_p) × √(m_p/m_α) = (2e/e) × √(m/4m) = 2 × 1/2 = 1:1.

Q2. [NEET 2021] Work done by magnetic force on a moving charge is:

Answer: Zero Magnetic force F = q(v × B) is always perpendicular to velocity v. Work = F · ds = F·v dt = 0 (since F ⊥ v). Magnetic force cannot change speed or kinetic energy of a charge.

Q3. [NEET 2021] A circular coil of 25 turns and radius 12 cm carries current of 1 A. What is B at centre?

Answer B = Nμ₀I/(2R) = 25 × 4π × 10⁻⁷ × 1 / (2 × 0.12) = 25 × 4π × 10⁻⁷ / 0.24
= 100π × 10⁻⁷ / 0.24 ≈ 1.31 × 10⁻⁴ T

Q4. [NEET 2020] The time period of a charged particle completing one revolution in a magnetic field is:

(a) Directly proportional to speed (b) Inversely proportional to speed (c) Independent of speed (d) Proportional to mass²

Answer: (c) T = 2πm/(qB). It depends on mass (m), charge (q) and field (B) only. It is completely independent of speed and radius. This is the key principle of the cyclotron.

Q5. [NEET 2019] Two long parallel wires carry currents I₁ = 3A and I₂ = 6A in the same direction. They are 4 cm apart. Force per unit length:

Answer f = μ₀I₁I₂/(2πd) = (4π×10⁻⁷×3×6)/(2π×0.04) = (2×10⁻⁷×18)/0.04 = 36×10⁻⁷/0.04 = 9×10⁻⁵ N/m (attractive).

Q6. [NEET 2018] The sensitivity of a galvanometer can be increased by:

(a) Decreasing number of turns (b) Using a weaker magnetic field (c) Using a stronger spring (d) Increasing the area of coil

Answer: (d) Current sensitivity = NBA/k. To increase sensitivity: increase N (turns), B (field strength), A (area of coil), or decrease k (spring constant). Increasing A (area) directly increases sensitivity.

💡 Rapid Revision — Key Formulas

F = qvB sinθ | Work by B = 0 | r = mv/qB | T = 2πm/qB (no v dependence)
Biot-Savart: B at centre = μ₀NI/2R; Infinite wire = μ₀I/2πd
Ampere: ∮B·dl = μ₀I_enc; Solenoid = μ₀nI; Toroid = μ₀NI/2πr
F on conductor = BIL sinθ; Between wires = μ₀I₁I₂/2πd (same dir = attract)
Torque on loop: τ = NIAB sinθ; m = NIA
Cyclotron: ν_c = qB/2πm; KE_max = q²B²R²/2m; cannot accelerate neutrons
Ammeter: shunt S in parallel (low R) | Voltmeter: R in series (high R)
1 A defined: 2×10⁻⁷ N/m force between wires 1 m apart

Numericals - Moving Charges and Magnetism - Class 12

CLASS 12 PHYSICS | NCERT SOLUTIONS

Chapter 4 — Moving Charges and Magnetism

25 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions

📋 Note: Solutions follow NCERT Class 12 Physics Chapter 4. Formulas used: r = mv/qB, T = 2πm/qB, B(centre) = μ₀NI/2R, B(wire) = μ₀I/2πd, B(solenoid) = μ₀nI, f = μ₀I₁I₂/2πd.

📝 NCERT Exercise Questions (4.1 – 4.16)

3 MarksQ1 (Ex 4.1). A circular coil of 30 turns and radius 8 cm carries a current of 6 A. Find the magnetic field (B) at the centre of the coil.

✓ Solution
Given: N = 30, R = 8 cm = 0.08 m, I = 6 A

B = Nμ₀I / (2R)

B = 30 × (4π × 10⁻⁷) × 6 / (2 × 0.08)
B = 30 × 4π × 6 × 10⁻⁷ / 0.16
B = (720π × 10⁻⁷) / 0.16 = 4500π × 10⁻⁷
B = 1.41 × 10⁻³ T = 1.41 mT
Direction: perpendicular to the plane of the coil (right-hand thumb rule).

3 MarksQ2 (Ex 4.2). A long straight wire carries a current of 35 A. Find the magnitude of B at a point 20 cm away from the wire.

✓ Solution
Given: I = 35 A, d = 20 cm = 0.20 m

B = μ₀I / (2πd) = (4π × 10⁻⁷ × 35) / (2π × 0.20)

B = (2 × 10⁻⁷ × 35) / 0.20 = (70 × 10⁻⁷) / 0.20
B = 3.5 × 10⁻⁵ T = 35 µT
Direction: tangential (circular loops around wire), given by right-hand thumb rule.

3 MarksQ3 (Ex 4.3). A long straight wire in the horizontal plane carries current of 50 A from south to north. Find B at a point 2.5 m east of the wire. (Give direction.)

✓ Solution
Given: I = 50 A (south to north), d = 2.5 m (east)

B = μ₀I / (2πd) = (2 × 10⁻⁷ × 50) / 2.5

B = 100 × 10⁻⁷ / 2.5 = 4 × 10⁻⁶ T = 4 µT
Direction: Current is northward (+Y), point is east (+X). By right-hand thumb rule, B at a point to the east of the wire = B is directed vertically downward (–Z direction, i.e., vertically into the ground).

3 MarksQ4 (Ex 4.4). A horizontal overhead power line carries current of 90 A from east to west. Find B at a point 1.5 m below the line.

✓ Solution
Given: I = 90 A (east to west), d = 1.5 m (below wire)

B = μ₀I / (2πd) = (4π × 10⁻⁷ × 90) / (2π × 1.5)

B = (2 × 10⁻⁷ × 90) / 1.5 = 180 × 10⁻⁷ / 1.5 = 1.2 × 10⁻⁵ T = 12 µT
Direction: Current west (−X), point 1.5 m below (−Z). Using right-hand rule: B at point below the wire points towards south (−Y direction).

3 MarksQ5 (Ex 4.5). A circular coil of 100 turns and radius 8 cm is connected to a power supply of 12 V using a wire of resistance 6 Ω. Calculate B at the centre. (Resistance of coil = 0.4 Ω per turn)

✓ Solution
Given: N = 100, R = 8 cm = 0.08 m, V = 12 V
Total resistance of coil = 0.4 × 100 = 40 Ω. Wire resistance = 6 Ω.
Total circuit resistance = 40 + 6 = 46 Ω
I = V / R_total = 12 / 46 ≈ 0.261 A

B = Nμ₀I / (2R) = 100 × 4π × 10⁻⁷ × 0.261 / (2 × 0.08)

B = 100 × 4π × 0.261 × 10⁻⁷ / 0.16
B = (327.9π × 10⁻⁷) / 0.16 ≈ 6.44 × 10⁻⁴ T ≈ 0.64 mT

3 MarksQ6 (Ex 4.6). A straight wire carrying a current of 12 A is bent into a semicircular arc of radius 2 cm. Find B at the centre (bent in vertical plane).

✓ Solution
Given: I = 12 A, R = 2 cm = 0.02 m (semicircular arc)
B due to a semicircular arc = half that of a full circular loop:

B = μ₀I / (4R) = (4π × 10⁻⁷ × 12) / (4 × 0.02)

B = (4π × 12 × 10⁻⁷) / 0.08 = (48π × 10⁻⁷) / 0.08
B = 600π × 10⁻⁷ = 1884 × 10⁻⁷
B ≈ 1.88 × 10⁻⁴ T (directed perpendicular to the plane of the arc, by right-hand rule).

3 MarksQ7 (Ex 4.7). A square coil of side 10 cm, consisting of 20 turns, carries a current of 12 A. The coil is suspended vertically and the normal to the plane makes 30° with a uniform horizontal magnetic field of 0.80 T. Calculate the torque on the coil.

✓ Solution
Given: Side = 10 cm = 0.10 m, N = 20, I = 12 A, B = 0.80 T, θ = 30°
Area A = (0.10)² = 0.01 m²

τ = NIAB sinθ

τ = 20 × 12 × 0.01 × 0.80 × sin 30°
τ = 20 × 12 × 0.01 × 0.80 × 0.5
τ = 0.96 N·m

3 MarksQ8 (Ex 4.8). Two long and parallel straight wires A and B carrying currents of 8 A and 5 A in the same direction are separated by a distance of 4 cm. Estimate the force on a 10 cm section of wire A.

✓ Solution
Given: I_A = 8 A, I_B = 5 A, d = 4 cm = 0.04 m, L = 10 cm = 0.10 m
Force per unit length: f = μ₀I_AI_B/(2πd)

F = μ₀I_AI_BL / (2πd) = (4π×10⁻⁷×8×5×0.10) / (2π×0.04)

F = (2×10⁻⁷×8×5×0.10) / 0.04 = (8×10⁻⁷) / 0.04
F = 2 × 10⁻⁵ N (attractive, since currents are in same direction)

3 MarksQ9 (Ex 4.9). A solenoid of length 0.5 m has 100 turns and carries a current of 2 A. Find B inside. If there is an iron core with relative permeability μ_r = 900, find B with core.

✓ Solution
Given: L = 0.5 m, N = 100, I = 2 A, μ_r = 900
n = N/L = 100/0.5 = 200 turns/m
Without core:

B₀ = μ₀nI = 4π×10⁻⁷×200×2 = 1600π×10⁻⁷

B₀ = 5.03 × 10⁻⁴ T = 0.503 mT

With iron core:

B = μ_rμ₀nI = 900 × 5.03 × 10⁻⁴

B = 0.452 T ≈ 0.45 T

3 MarksQ10 (Ex 4.10). A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, find B (a) outside toroid (b) inside core (c) in empty space surrounded by toroid.

✓ Solution
Given: r_inner = 25 cm = 0.25 m, r_outer = 26 cm = 0.26 m, N = 3500, I = 11 A
Mean radius r = (0.25 + 0.26)/2 = 0.255 m

(a) B outside toroid = 0 (Ampere's law: net current enclosed = 0 outside a toroid)

(b) B inside core:

B = μ₀NI / (2πr) = (4π×10⁻⁷×3500×11) / (2π×0.255)

B = (2×10⁻⁷×3500×11) / 0.255 = (7700×10⁻⁷) / 0.255
B ≈ 3.02 × 10⁻³ T = 3.02 mT

(c) B in empty space inside hole = 0 (no current encloses that region)

3 MarksQ11 (Ex 4.11). A galvanometer coil has a resistance of 12 Ω and shows full-scale deflection for a current of 3 mA. Convert it into a voltmeter of range 0 to 18 V and an ammeter of range 0 to 6 A.

✓ Solution
Given: R_g = 12 Ω, I_g = 3 mA = 3 × 10⁻³ A

(a) Voltmeter (0–18 V):
Series resistance required:

R = V/I_g − R_g = 18/(3×10⁻³) − 12 = 6000 − 12

R = 5988 Ω ≈ 5988 Ω in series.

(b) Ammeter (0–6 A):
Shunt resistance required:

S = I_gR_g / (I − I_g) = (3×10⁻³×12) / (6 − 3×10⁻³)

S = 0.036 / 5.997 ≈ 0.036 / 6
S = 6 × 10⁻³ Ω = 6 mΩ in parallel.

3 MarksQ12 (Ex 4.12). A proton enters a magnetic field of 1.5 T with velocity 2 × 10⁷ m/s at a right angle to the field. Find the radius of circular orbit, frequency, and time for one revolution. (m_p = 1.67 × 10⁻²⁷ kg, e = 1.6 × 10⁻¹⁹ C)

✓ Solution
Given: B = 1.5 T, v = 2×10⁷ m/s, θ = 90°, m_p = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C

Radius:

r = mv/(qB) = (1.67×10⁻²⁷×2×10⁷) / (1.6×10⁻¹⁹×1.5)

r = (3.34×10⁻²⁰) / (2.4×10⁻¹⁹) = 0.139 m ≈ 14 cm

Time period:

T = 2πm/(qB) = (2π×1.67×10⁻²⁷) / (1.6×10⁻¹⁹×1.5)

T = (10.49×10⁻²⁷) / (2.4×10⁻¹⁹) = 4.37×10⁻⁸ s

Frequency: f = 1/T = 1/(4.37×10⁻⁸) ≈ 2.29×10⁷ Hz ≈ 22.9 MHz

3 MarksQ13 (Ex 4.13). An alpha particle (charge = 2e, mass = 6.64 × 10⁻²⁷ kg) is subjected to a magnetic field of 0.25 T perpendicular to its velocity of 3 × 10⁴ m/s. Find: (a) radius of circular path, (b) period of revolution.

✓ Solution
Given: q = 2e = 2×1.6×10⁻¹⁹ = 3.2×10⁻¹⁹ C, m = 6.64×10⁻²⁷ kg, B = 0.25 T, v = 3×10⁴ m/s

(a) Radius:

r = mv/(qB) = (6.64×10⁻²⁷×3×10⁴) / (3.2×10⁻¹⁹×0.25)

r = (19.92×10⁻²³) / (8×10⁻²⁰) = 2.49×10⁻³ m ≈ 2.49 mm

(b) Period:

T = 2πm/(qB) = (2π×6.64×10⁻²⁷) / (3.2×10⁻¹⁹×0.25)

T = (41.7×10⁻²⁷) / (8×10⁻²⁰) = 5.21×10⁻⁷ s

3 MarksQ14 (Ex 4.14). A rectangular loop of sides 25 cm and 10 cm carrying a current of 15 A is placed with its longer side parallel to a long straight wire 2 cm away carrying 25 A. Find the net force on the loop.

✓ Solution
Given: Wire current I₁ = 25 A; Loop current I₂ = 15 A; Loop sides = 25 cm (L) and 10 cm (w). Longer side parallel to wire at d = 2 cm = 0.02 m. Far side at d + w = 0.02 + 0.10 = 0.12 m.

Force on near side (attractive, same current direction):

F_near = μ₀I₁I₂L/(2πd₁) = (2×10⁻⁷×25×15×0.25)/0.02

F_near = (2×10⁻⁷×375×0.25)/0.02 = (18.75×10⁻⁶)/0.02 = 9.375×10⁻⁴ N (towards wire)

Force on far side (repulsive):

F_far = (2×10⁻⁷×25×15×0.25)/0.12 = (18.75×10⁻⁶)/0.12 = 1.563×10⁻⁴ N (away from wire)

Net force = F_near − F_far = 9.375×10⁻⁴ − 1.563×10⁻⁴ = 7.81 × 10⁻⁴ N towards the wire (attractive).

3 MarksQ15 (Ex 4.15). A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If current carried is 8 A, find B inside.

✓ Solution
Given: L = 80 cm = 0.80 m, layers = 5, turns per layer = 400, I = 8 A
Total turns: N = 5 × 400 = 2000
n = N/L = 2000/0.80 = 2500 turns/m

B = μ₀nI = 4π×10⁻⁷×2500×8

B = 4π×2000×10⁻⁷ = 8000π×10⁻⁷
B = 2.51×10⁻² T = 25.1 mT
(Diameter does not affect B inside a solenoid)

3 MarksQ16 (Ex 4.16). The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

✓ Solution
Given: I₁ = I₂ = 300 A (currents flow in opposite directions — battery wires always carry currents in opposite directions), d = 1.5 cm = 0.015 m

f = μ₀I₁I₂/(2πd) = (2×10⁻⁷×300×300)/0.015

f = (2×10⁻⁷×90000)/0.015 = (18×10⁻³)/0.015 = 1.2 N/m
Since currents are in opposite directions (one wire carries current from battery to motor, other returns from motor to battery), the force is repulsive.

🌟 Additional NCERT Exemplar / HOT Questions

3 MarksQ17. An electron (m = 9.11×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C) moves with velocity 2×10⁶ m/s in a magnetic field of 0.5 T perpendicular to its path. Find radius, period, and kinetic energy.

✓ Solution
r = mv/(eB) = (9.11×10⁻³¹×2×10⁶) / (1.6×10⁻¹⁹×0.5) = (1.822×10⁻²⁴) / (8×10⁻²⁰) = 2.28×10⁻⁵ m = 0.023 mm
T = 2πm/(eB) = (2π×9.11×10⁻³¹) / (1.6×10⁻¹⁹×0.5) = 7.13×10⁻¹¹ s
KE = ½mv² = ½×9.11×10⁻³¹×(2×10⁶)² = 1.82×10⁻¹⁸ J = 11.4 eV

3 MarksQ18. A circular loop of radius 0.1 m carries a current of 10 A placed in a uniform magnetic field of 0.5 T. Find the (a) maximum torque, (b) minimum torque, and (c) the position of stable equilibrium.

✓ Solution
N = 1, I = 10 A, R = 0.1 m, B = 0.5 T, A = πR² = π×0.01 = 0.0314 m²

(a) τ_max = NIAB sin 90° = 1×10×0.0314×0.5×1 = 0.157 N·m (plane of loop parallel to B)
(b) τ_min = NIAB sin 0° = 0 (plane of loop perpendicular to B)
(c) Stable equilibrium: magnetic moment m parallel to B (normal to plane along B direction). Potential energy = −mB (minimum).

3 MarksQ19. A galvanometer has G = 50 Ω and deflects full scale at I_g = 1 mA. It is to be converted to an ammeter for 5 A range. Find S. What fraction of total current flows through the galvanometer?

✓ Solution
R_g = 50 Ω, I_g = 1×10⁻³ A, I = 5 A

S = I_gR_g/(I−I_g) = (10⁻³×50)/(5−10⁻³) = 0.05/4.999 ≈ 0.01 Ω

Fraction through galvanometer = I_g/I = 10⁻³/5 = 2×10⁻⁴ (only 0.02% of current through galvanometer).

3 MarksQ20. In a cyclotron, protons are accelerated through a dee of radius R = 0.5 m with magnetic field B = 1.5 T. Find the maximum kinetic energy achieved. (m_p = 1.67×10⁻²⁷ kg)

✓ Solution
q = e = 1.6×10⁻¹⁹ C, B = 1.5 T, R = 0.5 m

KE_max = q²B²R² / (2m) = (1.6×10⁻¹⁹)²×(1.5)²×(0.5)² / (2×1.67×10⁻²⁷)

KE = (2.56×10⁻³⁸×2.25×0.25) / (3.34×10⁻²⁷)
KE = (1.44×10⁻³⁸) / (3.34×10⁻²⁷) = 4.31×10⁻¹² J
KE_max ≈ 4.31×10⁻¹² J ≈ 26.9 MeV

3 MarksQ21. Two concentric circular coils X and Y of radii 16 cm and 10 cm respectively lie in the same vertical plane. Coil X has 20 turns and 16 A current; Coil Y has 25 turns and 18 A (opposite direction). Find net B at common centre.

✓ Solution
B_X = N_Xμ₀I_X/(2R_X) = 20×4π×10⁻⁷×16/(2×0.16) = 20×4π×16×10⁻⁷/0.32 = (1280π×10⁻⁷)/0.32 = 4000π×10⁻⁷
B_X = 1.257×10⁻³ T

B_Y = N_Yμ₀I_Y/(2R_Y) = 25×4π×10⁻⁷×18/(2×0.10) = 25×4π×18×10⁻⁷/0.20 = (1800π×10⁻⁷)/0.20 = 9000π×10⁻⁷
B_Y = 2.827×10⁻³ T

Since currents are in opposite direction: B_net = B_Y − B_X = 2.827×10⁻³ − 1.257×10⁻³
B_net = 1.57×10⁻³ T = 1.57 mT (in direction of B_Y)

3 MarksQ22. A straight wire 0.35 m long carries a current of 28 A in a uniform magnetic field of 0.5 T making an angle of 60° with the wire. Find the magnitude of force on the wire.

✓ Solution
Given: L = 0.35 m, I = 28 A, B = 0.5 T, angle between wire and B = 60°
Since θ is angle between current direction and B, θ = 60°

F = BIL sinθ = 0.5×28×0.35×sin60°

F = 0.5×28×0.35×(√3/2) = 4.9×0.866 = 4.24 N

3 MarksQ23. A galvanometer of resistance 20 Ω has a full-scale deflection at 50 mA. How to convert it into a voltmeter reading up to 100 V?

✓ Solution
R_g = 20 Ω, I_g = 50 mA = 0.05 A, V = 100 V

R = V/I_g − R_g = 100/0.05 − 20 = 2000 − 20 = 1980 Ω

A 1980 Ω resistance in series is required. (Total resistance = 2000 Ω; when 100V applied, current = 100/2000 = 0.05 A = full-scale deflection.)

3 MarksQ24. A beam of protons (charge e, mass m_p) enters a region where E = 10⁴ V/m (upward) and B = 0.1 T (into the page). What speed v must protons have to pass undeflected? (Velocity selector)

✓ Solution
For a particle to pass undeflected, electric force must equal magnetic force:
qE = qvB

v = E/B = 10⁴ / 0.1 = 10⁵ m/s = 100 km/s

This is the principle of a velocity selector (Wien filter). Only particles with this exact speed pass straight through; others are deflected.

3 MarksQ25. State and prove that the time period of a charged particle moving in a magnetic field is independent of its velocity. Why is this property important for the cyclotron?

✓ Solution
Statement: The time period T of circular motion of a charged particle (charge q, mass m) in a magnetic field B is T = 2πm/(qB), which is independent of both velocity v and radius r.

Proof:
Magnetic force = centripetal force: qvB = mv²/r ⇒ r = mv/(qB)
T = circumference/speed = 2πr/v = 2π(mv/qB)/v = 2πm/(qB) — no v!

Importance for cyclotron: Since T is constant regardless of speed, a charged particle always completes each semicircle in the same time. The AC oscillator driving the dees is set at frequency f = 1/T = qB/(2πm). As the particle accelerates and gains speed (and larger radius), it still returns to the gap in sync with the oscillator, which now has the correct polarity to accelerate it again. This isochronous property is what makes the cyclotron work.

✍ Score Guide — 25 Questions
NCERT Ex 4.1–4.16: 3 marks each — total 48 marks
Additional HOT/Exemplar Q17–Q25: 3 marks each — total 27 marks
Grand Total: 75 marks

Formula Capsule - Moving Charges and Magnetism - Class 12

CLASS 12 PHYSICS | FORMULA CAPSULE

Moving Charges and Magnetism

Chapter 4 — Complete Formula Sheet & High-Yield Facts for NEET/JEE

📅 Section A — Magnetic Force on a Charge

Quantity / Formula	Expression	Key Condition / Note
Lorentz Force (magnetic)	F = qvB sinθ	θ = angle between v and B
Lorentz Force (total)	F = q(E + v × B)	Combined electric + magnetic
Work by magnetic force	W = 0	Always! F ⊥ v at every instant
F when θ = 0° or 180°	F = 0	Particle along or against B; no deflection
F when θ = 90°	F = qvB (maximum)	Most common in problems
Radius of circular motion	r = mv/(qB) = p/(qB)	Depends on momentum p = mv
Time period (in B)	T = 2πm/(qB)	Independent of v and r !
Cyclotron frequency	f = qB/(2πm)	Resonance condition for cyclotron
Angular velocity	ω = qB/m	Constant for given q, m, B

📅 Section B — Biot-Savart & Field Due to Currents

Configuration	Formula	Note
Biot-Savart element	dB = (μ₀/4π) I dl sinθ / r²	dB ⊥ both dl and r
Infinite straight wire	B = μ₀I / (2πd)	d = distance from wire
Semi-infinite wire	B = μ₀I / (4πd)	Half of infinite wire's field
Circular loop at centre (N turns)	B = Nμ₀I / (2R)	Most frequently tested
Circular loop at axial point x	B = μ₀IR² / [2(R²+x²)^3/2]	At x=0: gives B=μ₀I/2R ✓
Semicircular arc at centre	B = μ₀I / (4R)	Half of full circle
Long solenoid (inside)	B = μ₀nI	n = N/L = turns per metre
Solenoid end face	B = μ₀nI / 2	Half of inside value
Toroid (inside)	B = μ₀NI / (2πr)	r = mean radius
Toroid (outside / inside hole)	B = 0	Net current enclosed = 0

📅 Section C — Force, Torque, Galvanometer & Cyclotron

Quantity	Formula	Note
Force on conductor	F = BIL sinθ	Max when I ⊥ B, Zero when I ∥ B
Force/length between wires	f = μ₀I₁I₂ / (2πd)	Same direction: attract; Opposite: repel
Definition of 1 Ampere	f = 2×10⁻⁷ N/m (at d = 1m)	Two parallel wires, I₁=I₂=1A
Magnetic dipole moment	m = NIA	N=turns, I=current, A=area. Unit: A·m²
Torque on current loop	τ = NIAB sinθ = m × B	Max at θ=90°, Zero at θ=0°
Galvanometer deflection	θ = (NBA/k) I	k = spring constant; θ ∝ I (linear scale)
Current sensitivity	θ/I = NBA/k	Increase by: ↑N, ↑B, ↑A, ↓k
Shunt for ammeter	S = I_gR_g / (I−I_g)	Low resistance in parallel
Series R for voltmeter	R = V/I_g − R_g	High resistance in series
Cyclotron max KE	KE_max = q²B²R²/(2m)	R = radius of dee
Cyclotron max velocity	v_max = qBR/m	At rim of dee
Velocity selector	v = E/B	Condition: qE = qvB (undeflected)
Potential energy of dipole	U = −mB cosθ = −m·B	Min (stable) at θ=0°; Max (unstable) at θ=180°

🧠 Memory Tricks

“WORK = ZERO” — Magnetic Force Magnetic force is always perpendicular to velocity. Since W = F·ds and F⊥ds, work = 0. Speed unchanged; direction changes only. KE is constant in pure magnetic field.

“T has no v” — Cyclotron T = 2πm/qB — look carefully: no v, no r! This is isochronism. The frequency of the oscillator is fixed = qB/2πm. As particle speeds up, radius ↑ but T stays constant.

“Same Attract, Opposite Repel” — Wires Two parallel wires: currents in same direction → attract each other (like two rows of moving positive charges flowing together). Opposite direction → repel. (Opposite of what you might expect from charge rules!)

“Low Parallel, High Series” — MCG Ammeter needs low resistance shunt in parallel (so circuit current divides; most bypasses galvanometer). Voltmeter needs high resistance in series (so very little current flows; galvanometer reads voltage).

“r = p/qB” — Ratio Problems Same KE: r ∝ √(m/q). Same momentum: r ∝ 1/q. Same speed: r ∝ m/q.
Proton vs Deuteron (same v): r_d/r_p = m_d/m_p = 2.
Proton vs Alpha (same KE): r_α/r_p = √2 (since m_α=4m, q_α=2e).

“Solenoid vs Toroid” Solenoid: B=μ₀nI inside; B≈0 outside (open-ended, straight). Toroid: B=μ₀NI/2πr inside material; B=0 everywhere else (closed ring → complete field confinement).

🔢 Critical Values & Constants

μ₀ = 4π×10⁻⁷ T·m/A μ₀/4π = 10⁻⁷ T·m/A Work by magnetic force = 0 1A: force = 2×10⁻⁷ N/m at 1m Toroid outside: B = 0 Solenoid outside: B ≈ 0 T = 2πm/qB (no v, no r) Ideal ammeter R = 0 Ideal voltmeter R = ∞ KE_max = q²B²R²/2m (cyclotron)

❌ Common Mistakes to Avoid

Confusing Biot-Savart and Ampere's Law: Biot-Savart is general (any shape). Ampere's law is easier but requires symmetry. Use Ampere's for straight wire, solenoid, toroid.
Cyclotron cannot accelerate electrons: Electrons become relativistic at low energies → mass increases → cyclotron frequency changes → resonance fails.
Speed does NOT change in B field: Magnetic force always ⊥ v → no work → no KE change. In velocity selector, speed stays same; direction unaffected if qE=qvB.
r depends on momentum: If same KE → r∝√(m/q). If same speed → r∝m/q. If same momentum → r∝1/q. Don't mix these up.
Toroid vs Solenoid: For toroid, outside B = 0. For solenoid, at end-face B = μ₀nI/2 (not zero).
Direction of force on wire: F = I(L×B). Use Fleming's LHR. Force direction changes for negative current carriers (conventional current vs electron flow).

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