Practice question
Question
An electron moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field
of \( 0.7 \, \text{T} \). What is the radius of its path? (Mass = \( 9.1 \times 10^{-31} \, \text{kg}
\), charge = \( 1.6 \times 10^{-19} \, \text{C} \))
Explanation
**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. Radius r = (mv/qB) . r = (9.1 × 10⁻³¹ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.7) = (1.82 × 10⁻²⁴/1.12 × 10⁻¹⁹) = 1.625 × 10⁻⁵ m = 1.625 × 10⁻³ cm . Using F = q v B sinθ, F = I l B sinθ,
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.