Magnetic Fields & Magnetic Force — A/L Physics

How do I practise Magnetic Fields & Magnetic Force?

1 question with a worked solution

1. A proton of charge q = 1.6 × 10⁻¹⁹ C and mass m = 1.67 × 10⁻²⁷ kg enters perpendicularly into a uniform magnetic field B = 0.50 T with velocity v = 2.0 × 10⁶ m s⁻¹. What is the radius of the circular path traversed by the proton?
   1. A) 1.05 cm (0.0105 m)
   2. B) 8.35 cm (0.0835 m)
   3. C) 2.09 cm (0.0209 m)
   4. D) 4.18 cm (0.0418 m)
   Check my answer

   Correct answer: D) 4.18 cm (0.0418 m)

   The magnetic Lorentz force provides the centripetal force: q v B = m v² / r => r = m v / (q B). Substituting values: r = (1.67 × 10⁻²⁷ kg × 2.0 × 10⁶ m s⁻¹) / (1.6 × 10⁻¹⁹ C × 0.50 T) = (3.34 × 10⁻²¹) / (0.80 × 10⁻¹⁹) ≈ 0.04175 m ≈ 4.18 cm.

   Show worked solution

   Step 1: Formula: r = m·v / (q·B). Step 2: r = (1.67 × 10⁻²⁷ × 2.0 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.50) = 3.34 × 10⁻²¹ / 8.0 × 10⁻²⁰ = 0.04175 m ≈ 4.18 cm. Answer: 4.18 cm (0.0418 m).
   Modelled on: the A/L Physics examination pattern

Get this auto-marked

Next in the syllabus

Where does Magnetic Fields & Magnetic Force fit in the syllabus?

Unit 8.1 · Year 2 · Term 1