Magnetic Fields of Conductors & Solenoids — A/L Physics

How do I practise Magnetic Fields of Conductors & Solenoids?

1 question with a worked solution

1. A long air-core solenoid of length 0.50 m has 1000 turns of closely wound insulated wire. If a steady current of 2.0 A flows through the winding, what is the magnetic flux density B at the center of the solenoid? (μ₀ = 4π × 10⁻⁷ T m A⁻¹)
   1. A) 2.51 × 10⁻³ T
   2. B) 1.26 × 10⁻³ T
   3. C) 5.03 × 10⁻³ T (1.6π × 10⁻³ T)
   4. D) 1.01 × 10⁻² T
   Check my answer

   Correct answer: C) 5.03 × 10⁻³ T (1.6π × 10⁻³ T)

   The magnetic field inside a long solenoid is B = μ₀ n I, where n = N / L is the turn density. Here n = 1000 / 0.50 m = 2000 turns m⁻¹. Thus: B = (4π × 10⁻⁷ T m A⁻¹) × (2000 m⁻¹) × (2.0 A) = 1.6π × 10⁻³ T ≈ 5.03 × 10⁻³ T.

   Show worked solution

   Step 1: Calculate turn density n = N / L = 1000 / 0.50 = 2000 turns m⁻¹. Step 2: Formula: B = μ₀ · n · I. Step 3: B = (4π × 10⁻⁷) × 2000 × 2.0 = 16π × 10⁻⁴ T = 1.6π × 10⁻³ T ≈ 5.03 × 10⁻³ T. Answer: 5.03 × 10⁻³ T (1.6π × 10⁻³ T).
   Modelled on: the A/L Physics examination pattern

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Where does Magnetic Fields of Conductors & Solenoids fit in the syllabus?

Unit 8.2 · Year 2 · Term 1