Skip to content

Practice question

Question

An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C )

Options

Choose one · Correct answer highlighted

Explanation

Given: An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 8 × 10⁶¹.6 × 10⁻¹⁹ × 0.15 = frac7.28 × 10⁻²⁴².4 × 10⁻²⁰= 3.033 × 10⁻⁴ m approx 0.0303 cm . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Discussion

Comments

0 comments

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