Moving Charges and Magnetism — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Moving Charges and Magnetism" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Moving Charges and Magnetism" — 7 important questions with detailed answers for CBSE b…
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NCERT Solutions for Class 12 Physics — Moving Charges and Magnetism. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Moving Charges and Magnetism — All 7 Questions Solved
Q1. What is magnetic field? State its SI unit and symbol. Write expression for magnetic field due to long straight current-carrying wire.
Magnetic Field (B): Region around magnet or current-carrying conductor where magnetic force is experienced.
SI Unit: Tesla (T) or Weber/m² (Wb/m²)
Symbol: B (vector quantity)
Smaller unit: Gauss (G) where 1 T = 10⁴ G
Magnetic Field due to Long Straight Wire:
B = μ₀I/(2πr)
where μ₀ = 4π × 10⁻⁷ T⋅m/A (permeability of free space)
I = current in wire
r = perpendicular distance from wire
Direction: Given by right-hand rule
- Thumb points in direction of current
- Fingers curl in direction of magnetic field
Field pattern: Concentric circles around wire, strength decreases as 1/r
Q2. Define magnetic force on moving charge. An electron moves with velocity 10⁶ m/s perpendicular to magnetic field of 0.5 T. Calculate force on electron.
Magnetic Force on Moving Charge: Force on charged particle moving in magnetic field.
F = qvB sin θ = qv × B (vector form)
where q = charge
v = velocity
B = magnetic field
θ = angle between v and B
SI Unit: Newton (N)
Direction: Given by right-hand rule (for positive charge)
- Point fingers in direction of velocity
- Curl fingers toward B
- Thumb points in direction of force
For negative charge, reverse the direction
Given: v = 10⁶ m/s
B = 0.5 T
θ = 90° (perpendicular), sin 90° = 1
q = -e = -1.6 × 10⁻¹⁹ C (electron charge)
Step 1: F = |q|vB sin θ = (1.6 × 10⁻¹⁹) × (10⁶) × (0.5) × (1)
Step 2: F = 1.6 × 10⁻¹⁹ × 5 × 10⁵
Step 3: F = 8 × 10⁻¹⁴ N
Final Answer: Force = 8 × 10⁻¹⁴ N (perpendicular to both v and B)
Direction: Using left-hand rule for negative charge
Q3. What is Hall effect? Explain its significance and write Hall coefficient expression.
Hall Effect: Development of transverse electric field across a conductor carrying current in presence of perpendicular magnetic field.
Mechanism:
1. Current flows through conductor due to drift of charge carriers
2. Perpendicular magnetic field exerts force on moving charges
3. Charges accumulate on one side, creating electric field
4. This field creates Hall voltage (V_H) perpendicular to both current and field
Hall Voltage:
V_H = BR×I/d
where R_H = Hall coefficient
B = magnetic field
I = current
d = thickness of conductor
Hall Coefficient:
R_H = 1/(ne)
where n = charge carrier density
e = charge per carrier
Significance:
1. Determines type of charge carriers (sign of R_H indicates positive/negative carriers)
2. Measures charge carrier density
3. Used in Hall sensors for measuring magnetic fields
4. Determines conductivity and mobility of charge carriers
5. Important in semiconductor physics and device physics
Applications:
- Magnetic field sensors
- Current measurement
- Material characterization
- Determining carrier type in semiconductors
Q4. Derive expression for magnetic field at center of circular current loop of radius R.
Consider circular loop of radius R carrying current I. Find field at center O.
Take small element dl at distance R from center.
Magnetic field due to element dl:
dB = (μ₀I dl)/(4π R²) (using Biot-Savart law)
Direction of dB: Perpendicular to plane containing dl and position vector (using right-hand rule), pointing outward from loop
For circular symmetry, components perpendicular to axis cancel.
Only axial component survives.
For loop in horizontal plane, vertical component of dB:
dB_vertical = (μ₀I dl)/(4π R²)
Total field at center:
B = ∮ dB = ∮ (μ₀I dl)/(4π R²)
B = (μ₀I)/(4π R²) ∮ dl
B = (μ₀I)/(4π R²) × 2πR
B = μ₀I/(2R)
Final Answer: B = μ₀I/(2R)
where μ₀ = 4π × 10⁻⁷ T⋅m/A
Direction: Perpendicular to loop plane (given by right-hand rule)
For N loops:
B = Nμ₀I/(2R)
Q5. What is magnetic dipole? Define magnetic dipole moment and give expression.
Magnetic Dipole: Current-carrying loop behaving like a magnet with north and south poles.
Magnetic Dipole Moment (m): Measure of strength of magnetic dipole.
m = IA
where I = current
A = area of loop
SI Unit: A⋅m² (Ampere-meter squared)
Alternatively: J/T (Joule per Tesla)
Direction: Perpendicular to plane of loop, given by right-hand rule
- Curl fingers in direction of current
- Thumb points in direction of m
Vector form: m = IA (vector; magnitude IA, direction perpendicular to loop)
Properties:
1. Magnetic moment is additive for multiple loops: m_total = n × IA (for n loops)
2. Torque on dipole in field: τ = m × B = mB sin θ
3. Potential energy: U = -m⋅B = -mB cos θ
4. Field on axis of dipole: B = (μ₀/4π) × (2m/r³)
Examples:
- Electron orbital motion: m = -(e/(2m_e)) × L
- Electron spin: m_s ≈ 1 Bohr magneton
- Atomic nuclei: m = μ_N × I_nuclear
Q6. A rectangular coil ABCD with sides 10 cm and 5 cm carries current 2 A. It is placed in magnetic field 0.5 T perpendicular to plane of coil. Calculate: (a) Magnetic dipole moment (b) Torque if field makes angle 30° with normal to coil
Given: Length L = 10 cm = 0.1 m
Breadth B = 5 cm = 0.05 m
Current I = 2 A
Magnetic field H = 0.5 T
(a) Magnetic Dipole Moment:
Area A = L × B = 0.1 × 0.05 = 0.005 m²
m = IA = 2 × 0.005 = 0.01 A⋅m²
Final Answer (a): m = 0.01 A⋅m² or 10 mA⋅m²
(b) Torque when field makes 30° with normal:
When normal makes 30° with field, the angle between m and B is 30°
Torque: τ = mB sin θ = mB sin 30°
τ = 0.01 × 0.5 × sin 30°
τ = 0.01 × 0.5 × 0.5
τ = 0.0025 N⋅m
τ = 2.5 × 10⁻³ N⋅m
Final Answer (b): Torque = 2.5 × 10⁻³ N⋅m or 2.5 mN⋅m
Direction: Perpendicular to both m and B (given by right-hand rule)
Q7. State Ampere's circuital law. Use it to find magnetic field inside a solenoid.
Ampere's Circuital Law: Line integral of magnetic field around closed loop = μ₀ times enclosed current.
∮ B⋅dl = μ₀I_enclosed
where B = magnetic field
dl = infinitesimal displacement element
I_enclosed = net current enclosed by loop
μ₀ = permeability of free space
For Solenoid (infinitely long, ideal case):
Solenoid: Coil of wire wound in helical form, many turns close together.
Consider Amperian rectangular loop with sides parallel and perpendicular to solenoid axis.
Step 1: Inside solenoid, field is uniform and parallel to axis. Outside, field ≈ 0
Step 2: Choose rectangular loop with length l inside and outside
Step 3: Apply Ampere's law:
∮ B⋅dl = B⋅l (only inside contributes)
Step 4: Current enclosed = nIl where n = turns per unit length, I = current
Step 5: B⋅l = μ₀nIl
Step 6: B = μ₀nI = μ₀(N/L)I
where N = total turns, L = length of solenoid
Final Answer: B = μ₀nI inside solenoid (uniform field, independent of position)
Outside ideal solenoid: B = 0
At ends: B ≈ μ₀nI/2 (average of inside and outside values)
Moving Charges and Magnetism — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A proton enters a uniform magnetic field of 0.5 T with a velocity of 4 x 10^6 m/s at an angle of 30° with the field. What is the magnitude of the magnetic force experienced by the proton? (Charge of proton = 1.6 x 10^-19 C)
- A. 1.6 x 10^-13 N
- B. 3.2 x 10^-13 N
- C. 6.4 x 10^-13 N
- D. 0 N
Answer: B. 3.2 x 10^-13 N — The magnetic force on a charged particle is given by F = qvBsinθ. Substituting the given values: F = (1.6 x 10^-19 C) * (4 x 10^6 m/s) * (0.5 T) * sin(30°) = 3.2 x 10^-13 N.
MCQ 2. Which of the following statements is true regarding magnetic field lines?
- A. They always intersect each other.
- B. They originate from the south pole and end at the north pole outside the magnet.
- C. They form closed loops.
- D. The density of field lines is inversely proportional to the strength of the magnetic field.
Answer: C. They form closed loops. — Magnetic field lines always form closed loops, starting from the north pole and ending at the south pole outside the magnet, and continuing inside the magnet from south to north. They never intersect each other, and their density indicates the strength of the field.
MCQ 3. According to Ampere's Circuital Law, the line integral of the magnetic field B around any closed path is equal to:
- A. μ₀ times the total current passing through the surface enclosed by the path.
- B. μ₀ / 2π times the total current.
- C. zero, if no current is enclosed.
- D. ε₀ times the total charge enclosed.
Answer: A. μ₀ times the total current passing through the surface enclosed by the path. — Ampere's Circuital Law states that the line integral of the magnetic field B around any closed path is equal to μ₀ times the net current I threading through the loop: ∮ B⋅dl = μ₀I_enclosed.
MCQ 4. A current-carrying conductor of length L is placed in a uniform magnetic field B. The force on the conductor is maximum when the angle between the conductor and the magnetic field is:
- A. 0°
- B. 30°
- C. 45°
- D. 90°
Answer: D. 90° — The force on a current-carrying conductor in a magnetic field is given by F = I(L x B) or F = BILsinθ. The force is maximum when sinθ = 1, which means θ = 90°.
MCQ 5. A charged particle moves in a circular path in a uniform magnetic field. Which of the following quantities remains constant?
- A. Speed
- B. Velocity
- C. Momentum
- D. Kinetic Energy and Velocity
Answer: A. Speed — When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force, continuously changing the direction of the velocity but not its magnitude. Hence, speed and kinetic energy remain constant, while velocity and momentum change direction.
MCQ 6. A moving coil galvanometer is converted into an ammeter by connecting a:
- A. High resistance in series
- B. Low resistance in series
- C. High resistance in parallel
- D. Low resistance in parallel
Answer: D. Low resistance in parallel — To convert a galvanometer into an ammeter, a very low resistance, called a shunt resistance, is connected in parallel with the galvanometer. This allows most of the current to bypass the galvanometer, protecting it and extending its range.
MCQ 7. What is the principle behind the working of a cyclotron?
- A. Magnetic force on a stationary charge.
- B. Electric force on a moving charge.
- C. Magnetic field due to a current loop.
- D. Charged particles can be accelerated to high energies by making them cross the same electric field repeatedly under the influence of a strong magnetic field.
Answer: D. Charged particles can be accelerated to high energies by making them cross the same electric field repeatedly under the influence of a strong magnetic field. — A cyclotron accelerates charged particles to high energies using a strong magnetic field to keep them in a spiral path and an alternating electric field to provide acceleration each time they cross the gap between the 'dees'.
MCQ 8. The magnetic field at the center of a circular current loop of radius R carrying current I is given by:
- A. μ₀I / (2πR)
- B. μ₀I / (4πR)
- C. μ₀I / (2R)
- D. μ₀IR / 2
Answer: C. μ₀I / (2R) — Using Biot-Savart Law, the magnetic field at the center of a circular current loop of radius R carrying current I is B = μ₀I / (2R).
MCQ 9. Two long parallel wires carrying currents I1 and I2 in the same direction are separated by a distance d. The force per unit length between them is:
- A. Attractive, μ₀I1I2 / (2πd)
- B. Repulsive, μ₀I1I2 / (2πd)
- C. Attractive, μ₀I1I2 / d
- D. Repulsive, μ₀I1I2 / d
Answer: A. Attractive, μ₀I1I2 / (2πd) — Parallel currents in the same direction attract each other. The force per unit length between two long parallel current-carrying wires is F/L = μ₀I1I2 / (2πd).
MCQ 10. An electron enters a region where a uniform magnetic field B is perpendicular to its velocity v. What is the work done by the magnetic field on the electron?
- A. Positive
- B. Negative
- C. Zero
- D. Depends on the magnitude of B
Answer: C. Zero — The magnetic force is always perpendicular to the velocity of the charged particle (F = q(v x B)). Since work done W = F⋅d = Fdcosθ, and the angle θ between force and displacement (direction of velocity) is 90°, cos(90°) = 0. Thus, the work done by the magnetic force is always zero.
MCQ 11. A charged particle moves undeflected in a region with both electric field E and magnetic field B. If E and B are perpendicular to each other, what is the speed of the particle?
- A. E/B
- B. B/E
- C. EB
- D. √(E/B)
Answer: A. E/B — For the particle to move undeflected, the electric force (qE) must balance the magnetic force (qvB). Thus, qE = qvB, which gives v = E/B. This is the principle of a velocity selector.
MCQ 12. What is the SI unit of magnetic flux density (magnetic field strength)?
- A. Weber
- B. Tesla
- C. Ampere-meter
- D. Henry
Answer: B. Tesla — The SI unit of magnetic flux density or magnetic field strength is Tesla (T). Weber is the unit of magnetic flux, Ampere-meter is not a standard unit for this, and Henry is the unit of inductance.
MCQ 13. A current loop of area A and carrying current I is placed in a uniform magnetic field B. The magnitude of the torque acting on the loop is given by:
- A. IA / B
- B. IBA cosθ
- C. IBA sinθ
- D. B / (IA)
Answer: C. IBA sinθ — The torque acting on a current loop is given by τ = M x B, where M = IA is the magnetic dipole moment. The magnitude is τ = M B sinθ = IAB sinθ, where θ is the angle between the magnetic moment vector (normal to the loop's plane) and the magnetic field B.
MCQ 14. Which of the following is NOT a property of magnetic field lines?
- A. They originate from the North pole and terminate at the South pole (outside the magnet).
- B. They form continuous closed loops.
- C. The tangent at any point on a field line gives the direction of the magnetic field at that point.
- D. They pass through soft iron but not through air.
Answer: D. They pass through soft iron but not through air. — Magnetic field lines can pass through air as well as magnetic materials. The other three options are correct properties of magnetic field lines.
MCQ 15. A long solenoid has 'n' turns per unit length and carries a current 'I'. The magnetic field inside the solenoid, far from its ends, is given by:
- A. μ₀nI
- B. μ₀I / n
- C. μ₀nI / 2
- D. μ₀nI²
Answer: A. μ₀nI — According to Ampere's Circuital Law, the magnetic field inside a long solenoid is uniform and given by B = μ₀nI, where n is the number of turns per unit length and I is the current.
MCQ 16. A rectangular coil of 100 turns, area 0.01 m², and carrying a current of 2 A is placed in a uniform magnetic field of 0.1 T. If the plane of the coil is parallel to the magnetic field, what is the torque acting on the coil?
- A. 0 Nm
- B. 0.02 Nm
- C. 0.2 Nm
- D. 2 Nm
Answer: B. 0.02 Nm — The torque is given by τ = NIABsinθ. Here, N=100, I=2A, A=0.01m², B=0.1T. If the plane is parallel to the field, the angle between the area vector (normal to the plane) and the magnetic field is 90°. So, τ = 100 * 2 * 0.01 * 0.1 * sin(90°) = 0.02 Nm.
MCQ 17. In a moving coil galvanometer, the current sensitivity can be increased by:
- A. Decreasing the number of turns in the coil.
- B. Decreasing the area of the coil.
- C. Increasing the restoring torque constant (k) of the suspension wire.
- D. Increasing the magnetic field strength (B).
Answer: D. Increasing the magnetic field strength (B). — Current sensitivity (S_I) is defined as the deflection per unit current (θ/I = NAB/k). To increase sensitivity, N, A, or B should be increased, or k should be decreased.
MCQ 18. Which rule is used to determine the direction of the magnetic field produced by a current-carrying straight wire?
- A. Fleming's Left-Hand Rule
- B. Fleming's Right-Hand Rule
- C. Right-Hand Thumb Rule
- D. Lorentz Rule
Answer: C. Right-Hand Thumb Rule — The Right-Hand Thumb Rule (or Maxwell's corkscrew rule) is used to find the direction of the magnetic field around a current-carrying wire. Fleming's rules are for forces/induced current, and Lorentz rule refers to the force on a moving charge.
MCQ 19. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same kinetic energy. What is the ratio of their radii of curvature (R_p / R_α)? (Given: m_α = 4m_p, q_α = 2q_p)
- A. 1:1
- B. 1:2
- C. 2:1
- D. √2:1
Answer: A. 1:1 — The radius of curvature is R = mv/qB. Kinetic energy K = (1/2)mv². So, v = √(2K/m). Substituting v, R = (m/qB)√(2K/m) = (√(2Km))/qB. For proton (p) and alpha (α): R_p = (√(2Km_p))/q_pB and R_α = (√(2Km_α))/q_αB. Since K and B are same, R_p/R_α = (√(m_p)/q_p) / (√(m_α)/q_α). R_p/R_α = (√(m_p)/q_p) / (√(4m_p)/2q_p) = (√(m_p)/q_p) / (2√(m_p)/2q_p) = 1:1.
MCQ 20. A stationary charge in a magnetic field experiences:
- A. A force in the direction of the magnetic field.
- B. A force perpendicular to the magnetic field.
- C. No magnetic force.
- D. A force if the magnetic field is non-uniform.
Answer: C. No magnetic force. — The magnetic force on a charge is given by F = q(v x B). If the charge is stationary, its velocity v = 0, so the magnetic force F = 0. A stationary charge does not experience a magnetic force.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.