Class 9 Physics Chapter 4 Moving Charges and Magnetism: Important Questions & Answers

Moving Charges and Magnetism occupies a critical position in Class 9 Physics because it bridges fundamental electrostatics (Chapter 2) and introduces the concept that moving charges generate magnetic fields. The CBSE board pattern prioritizes questions on: • Biot-Savart law and its applications to straight wires and circular loops • Ampere's circuital law for calculating magnetic field strength • Direction of magnetic force (using Fleming's left-hand rule) on current-carrying conductors • Cyclotron motion and its practical importance in particle accelerators These topics carry approximately 8–12 marks in a standard Class 9 summative assessment. The board favours conceptual understanding over rote memorization, often asking candidates to apply formulas to real-world scenarios (e.g., calculating the magnetic field at the centre of a circular coil or determining the Lorentz force on an electron beam). By mastering the questions in this guide, you develop both procedural skill (solving numerical problems) and conceptual depth (understanding *why* moving charges create magnetic fields). This foundation is essential for Class 10 and Class 11 electromagnetism.

**Question 1:** The magnetic field due to a long straight current-carrying wire at a perpendicular distance r is inversely proportional to: (A) r (B) r² (C) r³ (D) √r **Answer:** (A) r **Explanation:** According to the Biot-Savart law, the magnetic field B at a point near a long straight wire carrying current I is given by B ∝ I/r. The field strength decreases linearly with distance from the wire. --- **Question 2:** Fleming's left-hand rule is used to determine the direction of: (A) Magnetic field around a current-carrying wire (B) Force on a current-carrying conductor in a magnetic field (C) Current in a solenoid (D) Induced EMF in a coil **Answer:** (B) Force on a current-carrying conductor in a magnetic field **Explanation:** Fleming's left-hand rule relates the direction of current (first finger), magnetic field (second finger), and resulting mechanical force (thumb) on a conductor. The right-hand rule, by contrast, gives the direction of the magnetic field around a current. --- **Question 3:** A charged particle moves perpendicular to a uniform magnetic field. The path traced is: (A) A straight line (B) A parabola (C) A circle (D) A helix **Answer:** (C) A circle **Explanation:** When a charged particle enters a uniform magnetic field perpendicularly, the magnetic force provides the centripetal force, causing circular motion. This principle is the basis for cyclotron operation. --- **Question 4:** The SI unit of magnetic field strength B is: (A) Tesla (T) (B) Weber (Wb) (C) Ampere (A) (D) Gauss (G) **Answer:** (A) Tesla (T) **Explanation:** 1 Tesla = 1 Newton per Ampere-metre (N·A⁻¹·m⁻¹) or 1 Weber per square metre (Wb·m⁻²). Gauss is a CGS unit, equal to 10⁻⁴ Tesla. --- **Question 5:** Ampere's circuital law states that the line integral of the magnetic field around a closed path is proportional to: (A) The enclosed charge (B) The enclosed current (C) The enclosed magnetic flux (D) The area enclosed **Answer:** (B) The enclosed current **Explanation:** Mathematically, ∮ B·dl = μ₀I_enclosed, where μ₀ is the permeability of free space. This is Ampere's law in its integral form and is analogous to Gauss's law in electrostatics.

**Question 1:** State Biot-Savart law and explain what each term represents. **Answer:** Biot-Savart law states that the magnetic field dB at a point P due to a small element dl of a current-carrying wire is: dB = (μ₀ / 4π) × (I × dl × sin θ) / r² Where: • dB = infinitesimal magnetic field at P • μ₀ = permeability of free space (4π × 10⁻⁷ T·m·A⁻¹) • I = current in the wire • dl = length of the current element • θ = angle between dl and the line joining dl to point P • r = perpendicular distance from dl to point P The law is fundamental for calculating magnetic fields for arbitrary current geometries. --- **Question 2:** A circular coil of radius R carries current I. Derive the magnetic field at the centre of the coil using Biot-Savart law. **Answer:** Consider a small element dl of the circular coil. For each element: • θ = 90° (dl ⊥ r), so sin θ = 1 • r = R (constant for a circle's circumference) Using Biot-Savart law: dB = (μ₀ / 4π) × (I·dl) / R² All field elements point in the same direction (perpendicular to the plane, by right-hand rule). Integrating around the entire circumference (∮ dl = 2πR): B = (μ₀ / 4π) × (I / R²) × 2πR = (μ₀·I) / (2R) For N turns: B = (μ₀·N·I) / (2R) --- **Question 3:** Two parallel wires carry currents I₁ and I₂ in the same direction, separated by distance d. Determine whether they attract or repel each other. **Answer:** Using Ampere's law, the magnetic field at distance d due to wire 1 is: B₁ = (μ₀·I₁) / (2πd) Wire 2 experiences a force F per unit length: F/L = B₁·I₂ = (μ₀·I₁·I₂) / (2πd) Using Fleming's left-hand rule: if currents flow in the same direction, the magnetic field from wire 1 at wire 2's location points perpendicular to I₂, causing an inward force. **Conclusion: The wires attract.** (Opposite currents repel.) --- **Question 4:** What is the Lorentz force? Write its vector form and explain when it is maximum. **Answer:** The Lorentz force F on a charge q moving with velocity v in a magnetic field B is: F = q(v × B) **Vector magnitude:** F = qvB sin θ, where θ is the angle between v and B. **Maximum force:** Occurs when sin θ = 1, i.e., θ = 90°. This means velocity is perpendicular to the magnetic field. F_max = qvB. **Zero force:** When v ∥ B (θ = 0°), the particle is undeflected and continues in a straight line. --- **Question 5:** Explain the working principle of a cyclotron and state one limitation. **Answer:** A cyclotron accelerates charged particles in a spiral path using a uniform magnetic field and alternating electric field. A particle with charge q and mass m, in a perpendicular magnetic field B, experiences a centripetal force: qvB = mv²/r This gives a circular path with radius r = mv/(qB). As the particle spirals outward, its kinetic energy increases through AC electric field pulses. The frequency of oscillation (cyclotron frequency) is f = qB/(2πm), which is **independent of velocity**—allowing synchronized acceleration. **Limitation:** Relativistic effects. As velocity approaches the speed of light, the particle's mass effectively increases, changing the cyclotron frequency and desynchronizing acceleration. Modern accelerators (synchrocyclotrons) adjust the frequency to compensate.

**Question 1:** A long solenoid with n turns per unit length carries current I. Using Ampere's law, derive the magnetic field inside the solenoid. Is the field uniform? **Answer:** Consider a rectangular Amperian loop partially inside and outside the solenoid. The loop has length L parallel to the solenoid axis. Applying Ampere's law: ∮ B·dl = μ₀I_enclosed • Inside the solenoid: B is parallel to the axis; ∮ B·dl = B·L • Outside the solenoid: B ≈ 0 • Enclosed current: I_enclosed = n·L·I (n turns per unit length, each carrying current I) Therefore: B·L = μ₀·n·L·I **B = μ₀·n·I** **Uniformity:** Yes, the field is approximately uniform inside an ideal solenoid (far from the ends) and negligible outside. Real solenoids show edge effects near the ends. --- **Question 2:** An electron (q = –1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg) enters a uniform magnetic field of 2 T perpendicularly with speed 3 × 10⁶ m·s⁻¹. Calculate the radius of its circular path and the period of one revolution. **Answer:** **Radius:** From qvB = mv²/r: r = mv / (qB) r = (9.1 × 10⁻³¹ × 3 × 10⁶) / (1.6 × 10⁻¹⁹ × 2) r = (2.73 × 10⁻²⁴) / (3.2 × 10⁻¹⁹) **r ≈ 8.5 × 10⁻⁶ m = 8.5 μm** **Period:** T = 2πr / v = 2πm / (qB) T = (2π × 9.1 × 10⁻³¹) / (1.6 × 10⁻¹⁹ × 2) T = (5.71 × 10⁻³⁰) / (3.2 × 10⁻¹⁹) **T ≈ 1.78 × 10⁻¹¹ s ≈ 17.8 ps** --- **Question 3:** A rectangular current loop (sides a and b) lies in a uniform magnetic field B. The plane of the loop makes an angle θ with the magnetic field. Explain why the net force on the loop is zero, but a torque acts on it. Derive the torque. **Answer:** **Why net force is zero:** Each side of the loop experiences a force F = BIℓ, where ℓ is the length. Opposite sides carry current in opposite directions; hence their forces are equal and opposite, canceling each other. **Why torque acts:** The perpendicular distances of opposite force pairs from the axis differ. Consider the two sides perpendicular to B (length a): • Force on each: F = BIa • Perpendicular distance between forces: b·sin θ • Torque: τ = F × (b·sin θ) = BIa·b·sin θ = BIA·sin θ Where A = a·b is the area of the loop. **In vector form:** τ = m × B, where magnetic moment m = IA (pointing perpendicular to the loop plane). This torque tends to align the loop plane perpendicular to B—fundamental to electric motors. --- **Question 4:** Two parallel, long, straight wires are separated by distance d and carry currents I₁ = 10 A and I₂ = 5 A in opposite directions. Calculate the magnetic field at the midpoint between them (μ₀ = 4π × 10⁻⁷ T·m·A⁻¹). **Answer:** At the midpoint, the distance from each wire is d/2. **Magnetic field due to wire 1:** B₁ = (μ₀·I₁) / (2π × d/2) = (μ₀·I₁) / (πd) B₁ = (4π × 10⁻⁷ × 10) / (π × d) = (4 × 10⁻⁶) / d **Magnetic field due to wire 2:** B₂ = (μ₀·I₂) / (πd) = (4π × 10⁻⁷ × 5) / (πd) = (2 × 10⁻⁶) / d **Direction:** Using the right-hand rule, with opposite currents, B₁ and B₂ point in the *same direction* at the midpoint. **Net field:** B_net = B₁ + B₂ = (4 × 10⁻⁶ + 2 × 10⁻⁶) / d = (6 × 10⁻⁶) / d Tesla For d = 1 m: **B_net = 6 × 10⁻⁶ T = 6 μT**

**Question 1:** State Ampere's circuital law. Derive an expression for the magnetic field due to a long straight current-carrying wire using this law. Compare with the Biot-Savart law result. **Solution:** **Ampere's Circuital Law:** The line integral of magnetic field B around any closed path equals μ₀ times the net current enclosed: ∮ B·dl = μ₀I_enclosed **Derivation for a long straight wire:** Consider a circular Amperian loop of radius r, concentric with the wire, with current I flowing along the wire's axis. 1. By symmetry, B is constant at distance r and tangent to the circle. 2. ∮ B·dl = B × 2πr (since B ∥ dl) 3. Enclosed current: I_enclosed = I 4. Applying Ampere's law: B × 2πr = μ₀I **Therefore: B = (μ₀I) / (2πr)** **Comparison with Biot-Savart law result:** Using Biot-Savart law, integrating over the entire wire (angle subtended: 0 to 2π): B = (μ₀I / 4π) × ∫(sin θ / r²) dl = (μ₀I / 4π) × (2 / r) = **(μ₀I) / (2πr)** **Conclusion:** Both laws yield identical results, confirming the internal consistency of magnetism theory. Ampere's law is more convenient for problems with high symmetry. --- **Question 2:** A charged particle (mass m, charge q) enters a uniform magnetic field B with velocity v perpendicular to the field. Derive the radius and period of circular motion. Explain the application in a mass spectrometer. **Solution:** **Derivation of radius:** The magnetic Lorentz force provides centripetal force: qvB = mv²/r **r = mv / (qB)** This radius is called the gyroradius or Larmor radius. **Derivation of period:** The circumference of the circular path is 2πr. The period T is: T = (2πr) / v = (2π × mv/qB) / v = 2πm / (qB) **T = (2πm) / (qB)** Note: T is **independent of velocity**, allowing particles of different speeds to exit simultaneously—essential for cyclotron operation. **Frequency (cyclotron frequency):** f = 1/T = (qB) / (2πm) **Application in Mass Spectrometry:** In a mass spectrometer: 1. Particles with the same charge q and velocity v but different masses m enter the magnetic field. 2. Each traces a circular path with r = mv / (qB). 3. Particles with larger mass have larger radius and land at different positions on a detector. 4. By measuring r (or deflection distance), we determine m precisely. 5. This technique separates isotopes and identifies unknown molecules. **Example:** Carbon-12 and Carbon-14 ions (same q, same v) have r₁₂/r₁₄ = m₁₂/m₁₄ = 12/14 ≈ 0.857, allowing clear separation. --- **Question 3:** A rectangular coil (length L, width W, n turns) rotates in a uniform magnetic field B with angular velocity ω. Derive the expression for induced EMF and explain the principle of an AC generator. If L = 0.2 m, W = 0.1 m, n = 50, B = 0.5 T, and ω = 100 rad·s⁻¹, calculate the peak EMF. **Solution:** **Derivation of induced EMF:** Magnetic flux through the coil at angle θ(t) = ωt: Φ = nBA cos(ωt) By Faraday's law: ε = –dΦ/dt = nBA × ω × sin(ωt) **ε = ε₀ sin(ωt), where ε₀ = nBAω (peak EMF)** **Principle of AC generator:** As the coil rotates: 1. The component of B perpendicular to the coil plane varies as cos(ωt). 2. Magnetic flux oscillates between +NBA and –NBA. 3. Rate of flux change (and hence induced EMF) follows a sinusoidal pattern. 4. This oscillating EMF drives alternating current through an external circuit. 5. The frequency of oscillation is f = ω/(2π) Hz (50 Hz in many countries). **Calculation:** ε₀ = nBAω A = L × W = 0.2 × 0.1 = 0.02 m² ε₀ = 50 × 0.5 × 0.02 × 100 ε₀ = 50 × 1 **ε₀ = 50 V** The RMS value of the AC EMF: ε_rms = ε₀ / √2 ≈ 35.4 V

**Question: Designing a Particle Accelerator for Medical Imaging** A hospital plans to install a cyclotron to accelerate protons for cancer radiotherapy. The design requires: • Proton mass: m_p = 1.67 × 10⁻²⁷ kg • Proton charge: q_p = 1.6 × 10⁻¹⁹ C • Magnetic field strength: B = 1.5 T • Target final kinetic energy: 70 MeV (1 MeV = 1.6 × 10⁻¹³ J) **Sub-questions:** (a) Calculate the cyclotron frequency (in MHz) for protons in this magnetic field. Explain why this frequency must remain constant during acceleration. (b) A proton is accelerated from rest. Using energy considerations, find its velocity when it reaches 70 MeV kinetic energy. (Ignore relativistic effects; use non-relativistic kinetic energy.) (c) Determine the radius of the circular path when the proton exits the cyclotron with the calculated velocity. (d) Discuss one practical limitation of real cyclotrons at high proton energies and suggest how modern accelerators overcome it. **Step-by-Step Solution:** **Step 1 (Part a):** Cyclotron frequency: f = (q_p × B) / (2π × m_p) f = (1.6 × 10⁻¹⁹ × 1.5) / (2π × 1.67 × 10⁻²⁷) f = (2.4 × 10⁻¹⁹) / (1.05 × 10⁻²⁶) f ≈ 2.29 × 10⁷ Hz = **22.9 MHz** **Why constant?** The period T = 2πm / (qB) is independent of velocity. All protons, regardless of energy, take the same time to complete a semicircle (in each "dee" of the cyclotron). The RF voltage frequency must match this to ensure protons always encounter an accelerating electric field in phase. **Step 2 (Part b):** Kinetic energy: KE = 70 MeV = 70 × 1.6 × 10⁻¹³ J = 1.12 × 10⁻¹¹ J From KE = ½m_p v²: v = √(2 × KE / m_p) v = √(2 × 1.12 × 10⁻¹¹ / 1.67 × 10⁻²⁷) v = √(1.34 × 10¹⁶) v ≈ **1.16 × 10⁸ m·s⁻¹** (about 39% the speed of light) **Step 3 (Part c):** Radius: r = m_p × v / (q_p × B) r = (1.67 × 10⁻²⁷ × 1.16 × 10⁸) / (1.6 × 10⁻¹⁹ × 1.5) r = (1.94 × 10⁻¹⁹) / (2.4 × 10⁻¹⁹) r ≈ **0.81 m = 81 cm** This radius determines the physical size of the cyclotron's magnet poles. **Step 4 (Part d):** **Limitation:** At very high velocities (approaching relativistic speeds), the proton's inertial mass effectively increases due to relativistic effects: m_eff ≈ γm, where γ = 1/√(1 – v²/c²). This changes the cyclotron frequency, desynchronizing protons from the RF voltage. Acceleration becomes inefficient or impossible. **Modern solution (Synchrocyclotron):** Adjust the RF frequency dynamically as proton energy increases, compensating for relativistic mass change. Alternatively, use a *synchrotron*, which increases both the RF frequency and magnetic field strength simultaneously, keeping the orbit radius constant while accelerating particles to higher energies (up to TeV range in research facilities). Medical cyclotrons typically operate below 100 MeV, where relativistic effects are manageable with careful design.

The 18 questions curated above represent the exact *types and difficulty levels* tested in Class 9 Physics board exams and periodic assessments. However, understanding a solution and *solving independently under time pressure* are very different skills. At **cbsetutor.ai**, our AI tutor personalizes your practice workflow: • **Daily adaptive drills:** The AI identifies which question types you struggle with (e.g., Ampere's law derivations vs. numerical problems) and generates fresh, similar questions at increasing difficulty. • **Real-time feedback on working:** When you submit a solution, the AI checks not just your final answer but your intermediate steps, algebraic reasoning, and diagram sketches—giving you pinpointed corrections before misconceptions solidify. • **Timed mock tests:** Simulate the board exam environment. The system tracks your speed, accuracy, and conceptual gaps across 1-mark MCQs, 2-mark shorts, 3-mark applied problems, and 5-mark long-answers—exactly mirroring the CBSE pattern. • **Offline flashcards & formula sheets:** Memorize key results (B = μ₀I/(2πr), r = mv/qB, τ = BIA sin θ) with spaced-repetition scheduling. • **Live doubt sessions:** Chat with expert tutors for Chapter 4 concepts you're unsure about—often resolved in 5–10 minutes rather than waiting for school. **Start a 3-day free trial at cbsetutor.ai** to unlock unlimited practice on Moving Charges and Magnetism (and all other Class 9 Physics chapters). No credit card required.