Chapter 4 – Moving Charges and Magnetism (3 Marks Questions with Answers)
Q1. Briefly describe Oersted’s experiment leading to the discovery of magnetic effect of current.
Answer: In 1820, Hans Christian Oersted discovered that an electric current produces a magnetic field. He placed a magnetic needle near a current-carrying wire and observed that the needle deflected whenever current passed through the wire. The direction of deflection depended on the direction of current. This showed that moving charges (electric current) produce a magnetic field.
Q2. On what factors does the force experienced by a charged particle moving in a magnetic field depend?
Answer: The magnetic force on a charged particle depends on:
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The magnitude of the charge (q).
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The velocity (v) of the charged particle.
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The strength of the magnetic field (B).
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The angle (θ) between the velocity and magnetic field.
Mathematically,
F = qvB sinθ
Q3. Write the three features observed at the interaction of a charged particle in the presence of both the electric field and the magnetic field.
Answer:
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The charged particle experiences both electric and magnetic forces simultaneously.
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The net force acting on the particle is the vector sum of electric and magnetic forces.
F = q(E + v × B) -
If electric and magnetic forces are equal and opposite, the particle moves undeflected. This principle is used in velocity selector devices.
Q4. Derive the expression for the force acting on a conductor carrying current in a uniform magnetic field.
Answer:
Consider a conductor of length L carrying current I placed in magnetic field B.
Each charge q moving with drift velocity v experiences force F = q(v × B).
Total force on the conductor:
F = BIL sinθ
where θ is the angle between current and magnetic field.
This force acts perpendicular to both I and B.
Q5. What is the nature of the trajectory of a charged particle in a uniform magnetic field when its initial velocity makes angles (i) 0°, (ii) 90°, and (iii) in between 0° and 90° with the direction of the magnetic field?
Answer:
(i) 0°: The particle moves in a straight line parallel to the field (no magnetic force).
(ii) 90°: The particle moves in a circular path perpendicular to B.
(iii) Between 0° and 90°: The path becomes helical (combination of circular and linear motion).
Q6. What is the pitch of the helical path traversed by a charged particle moving in a uniform magnetic field? Obtain an expression for it.
Answer:
The pitch is the linear distance covered by the particle along the magnetic field in one revolution.
Let vₗ be the velocity component parallel to B and T be the time period.
Then,
Pitch = vₗ × T
and
T = (2πm) / (qB)
Hence,
Pitch = (2πm vₗ) / (qB)
Q7. State and explain Biot–Savart’s law. Also calculate the magnetic field at a point due to a small current element.
Answer:
Biot–Savart’s law states that the magnetic field dB due to a small current element Idl at a point is proportional to current I, length element dl, and sine of the angle θ between dl and the line joining element to point, and inversely proportional to square of distance r.
dB = (μ₀ / 4π) × (I dl sinθ) / r²
For a small element of 1 cm, I = 2 A, and r = 1 m,
B = (2 × 10⁻⁹ T), directed perpendicular outward.
Q8. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field. Calculate the magnitude of the magnetic field.
Answer:
Given: m = 0.2 kg, L = 1.5 m, I = 2 A,
For equilibrium: BIL = mg
∴ B = mg / (IL)
B = (0.2 × 9.8) / (2 × 1.5) = 0.65 T
Q9. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Answer:
Magnetic field due to a long straight conductor:
B = (μ₀ I) / (2πr)
Substitute: μ₀ = 4π × 10⁻⁷ H/m, I = 90 A, r = 1.5 m
B = (4π × 10⁻⁷ × 90) / (2π × 1.5) = 1.2 × 10⁻⁵ T
Direction: Towards South (by Right-Hand Thumb Rule).
Chapter 5 – Magnetism and Matter (3 Marks Questions with Answers)
Q1. Mention any three properties of magnetic field lines.
Answer:
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Magnetic field lines form continuous closed loops.
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They never intersect each other.
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The direction of the field line gives the direction of the magnetic field (from North to South outside the magnet).
Q2. State and explain Gauss’s law in magnetism. Write its significance.
Answer:
Gauss’s law in magnetism states that the total magnetic flux through any closed surface is zero:
∮ B · dA = 0
This implies that magnetic monopoles do not exist — magnetic field lines always form closed loops (start and end at poles of the same magnet).
Q3. Derive the expression for magnetic potential energy of a dipole placed in a uniform magnetic field.
Answer:
If a magnetic dipole of moment M is placed in a uniform magnetic field B making an angle θ,
Potential energy,
U = – M B cosθ
It is minimum (–MB) when the dipole aligns with the field, and maximum (+MB) when opposite.
Q4. List any three properties of diamagnetic materials.
Answer:
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Weakly repelled by magnetic fields.
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Magnetic susceptibility (χ) is small and negative.
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Independent of temperature (e.g., Bismuth, Copper).
Q5. Write any three properties of paramagnetic materials.
Answer:
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Weakly attracted by magnetic fields.
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Magnetic susceptibility (χ) is small and positive.
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χ decreases with increase in temperature (follows Curie’s Law).
Q6. Mention any three properties of ferromagnetic materials.
Answer:
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Strongly attracted by magnetic fields.
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High and positive magnetic susceptibility.
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Exhibit hysteresis (retain magnetism even after the field is removed).
Q7. A solenoid has a core of material with relative permeability 400. Windings carry a current of 2 A with 1000 turns per metre. Calculate (a) Magnetic intensity (H), (b) Magnetization (M), and (c) Magnetic field (B).
Answer:
Given: μᵣ = 400, I = 2 A, n = 1000 turns/m
(a) H = nI = 1000 × 2 = 2 × 10³ A/m
(b) M = (μᵣ – 1)H = 399 × 2 × 10³ = 7.98 × 10⁵ A/m ≈ 8 × 10⁵ A/m
(c) B = μ₀μᵣH = 4π × 10⁻⁷ × 400 × 2 × 10³ = 1.0 T
Q8. A bar magnet of magnetic moment 1.5 J/T lies along a uniform magnetic field of 0.22 T. Find the work required to turn it (i) normal, (ii) opposite to the field.
Answer:
Work done in rotating a dipole: W = MB (cosθ₁ – cosθ₂)
(i) From 0° to 90°: W = 1.5 × 0.22 (1 – 0) = 0.33 J
(ii) From 0° to 180°: W = 1.5 × 0.22 (1 – (–1)) = 0.66 J
