Magnetism and Matter
Punjab Board · Class 12 · Physics
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A bar magnet of magnetic moment 2.5 A·m² is placed in a uniform magnetic field of 0.2 T such that its axis makes an angle of 30° with the field. What is the torque acting on the magnet and the potential energy of the system?
The axial magnetic field of a short bar magnet at a distance of 20 cm from its centre is 0.16 T. What is the equatorial field at the same distance?
A solenoid has 1500 turns per metre and carries a current of 3 A. Its core has a relative permeability of 500. What is the magnetisation M of the core material?
A bar magnet of magnetic moment 5 A·m² and length 10 cm is placed with its north pole pointing north. At what distance on its equatorial line will the net magnetic field be zero, given that the horizontal component of Earth's field is 4 × 10⁻⁵ T? (Take μ₀/4π = 10⁻⁷ T·m/A)
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Which of the following correctly explains why Gauss's law for magnetism states that the net magnetic flux through any closed surface is always zero?
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Magnetic monopoles do not exist; magnetic field lines always form closed loops so equal flux enters and exits any closed surface.
Step 1: Gauss's law for magnetism: ΣB·ΔS = 0 over any closed surface. Step 2: This is a direct consequence of the non-existence of magnetic monopoles — there are no isolated sources or sinks of magnetic field. Step 3: Since magnetic field lines form continuous closed loops, every line that enters a closed surface must exit it, giving zero net flux. Step 4: Option B is wrong — the field inside can be non-zero. Option C contradicts the fact that magnetic field lines don't start or end (unlike electric field lines). Option D is a meaningless statement — flux is a scalar but not always zero for al
A solenoid with 2000 turns per metre carries a current of 1.5 A and has a core with relative permeability 600. Calculate the magnetic field B inside the solenoid. (μ₀ = 4π × 10⁻⁷ T·m/A)
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2.26 T
Step 1: First calculate H = nI = 2000 × 1.5 = 3000 A/m. Step 2: Use B = μ₀μᵣH. Step 3: B = 4π × 10⁻⁷ × 600 × 3000. Step 4: B = 4π × 10⁻⁷ × 1.8 × 10⁶ = 4π × 0.18 = 4 × 3.1416 × 0.18 ≈ 2.26 T. Step 5: Option B ignores μᵣ (uses only μ₀). Option C halves the correct answer — a common arithmetic slip. Option D incorrectly doubles. Always use B = μ₀μᵣH when a magnetic core is present.
A magnetic needle of moment m is oscillating in a uniform magnetic field B. If the time period of oscillation is T, and the moment of inertia of the needle is I, which expression correctly gives the time period?
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T = 2π√(I/mB)
Step 1: The restoring torque on a magnetic needle in a field is τ = −mB sinθ ≈ −mBθ for small θ. Step 2: This is analogous to simple harmonic motion where the restoring constant is mB. Step 3: For angular SHM: T = 2π√(I/k_restoring) = 2π√(I/mB). Step 4: Option B inverts I and mB — a common confusion with the formula. Option C misses the factor of 2. Option D omits m — the magnetic moment is crucial in determining the restoring torque. This formula is used to experimentally determine m or B.
Two short bar magnets A and B have magnetic moments 1.2 A·m² and 1.0 A·m² respectively. They are placed on the same horizontal surface with their axes along the same straight line with like poles facing each other, separated by a distance of 20 cm. At which point between them (measured from A) is th
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≈ 10.5 cm from A
Step 1: Let the neutral point be at distance x from A. Then distance from B = (0.20 − x). Step 2: On the axial line between like poles, both magnets produce fields in opposite directions. For zero net field: B_A = B_B. Step 3: (μ₀/4π)(2m_A/x³) = (μ₀/4π)(2m_B/(0.20−x)³). Step 4: m_A/x³ = m_B/(0.20−x)³ → 1.2(0.20−x)³ = 1.0·x³. Step 5: Taking cube root ratio: (0.20−x)/x = (1/1.2)^(1/3) = (0.833)^(1/3) ≈ 0.941. So 0.20 = x(1 + 0.941) = 1.941x → x ≈ 0.103 m ≈ 10.3 cm ≈ 10.5 cm. Option B assumes equal moments. Options C and D are off due to calculation errors.
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