Magnetism and Matter
Punjab Board · Class 12 · Physics
Most important questions from Magnetism and Matter for Punjab Board Class 12 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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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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