Question Detail
Question
A metal ball of 10kg with 5 C charge is moving perpendicular to a magnetic field of 5T. The radius of the path is 80 m. What is the speed of the ball and the period of the ball?
Correct Answer
Option D is the correct answer.
Detailed Explanation
When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force, causing it to move in a circular path.Magnetic Force (F_B) = qvBCentripetal Force (F_c) = mv²/rEquating these: qvB = mv²/rSolving for speed (v): v = qBr / mGiven: m = 10 kg, q = 5 C, B = 5 T, r = 80 m.v = (5 C * 5 T * 80 m) / 10 kg = (25 * 80) / 10 = 2000 / 10 = 200 m/s.The period (T) of circular motion is the circumference divided by speed:T = 2πr / vT = (2π * 80 m) / 200 m/s = 160π / 200 s = 0.8π s.T ≈ 0.8 * 3.14159 ≈ 2.51 s.So, the speed is 200 m/s and the period is approximately 2.5 s.
Hint
Equate the magnetic force (qvB) to the centripetal force (mv²/r) to find speed. Then use the relationship between speed, radius, and period (T = 2πr/v).