Practice question

Given: A proton moves at 1.5 × 10⁷ m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 1.67 × 10⁻²⁷ 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 = frac1.67 × 10⁻²⁷ × 1.5 × 10⁷¹.6 × 10⁻¹⁹ × 0.2 = frac2.505 × 10⁻²⁰³.2 × 10⁻²⁰= 0.7828 approx 0.78 m . 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.