Practice question

Given: A proton moves with a speed of 1 × 10⁶ m/s perpendicular to a magnetic field of 0.8 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: Radius r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac1.67 × 10⁻²⁷ × 1 × 10⁶¹.6 × 10⁻¹⁹ × 0.8 = frac1.67 × 10⁻²¹¹.28 × 10⁻¹⁹= 1.304 × 10⁻² m = 1.3 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.