Skip to content

Question

An electron moves at \( 2.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.5 \, \text{T}
\). What is the radius of its path? (Mass = \( 9.1 \times 10^{-31} \, \text{kg} \), charge = \( 1.6
\times 10^{-19} \, \text{C} \))

Options

Choose one · Correct answer highlighted

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. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 2.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (2.275 × 10⁻²⁴/8 × 10⁻²⁰) = 2.84375 × 10⁻⁵ m ≈ 2.84 × 10⁻³ cm . Using F = q v B sinθ, F = I l B sinθ, B

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.