A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius 0.20 m at a speed of 2.0 m/s. what is the magnitude of the charge on the particle

0.01 C

Explanation:

Applying,

F = qvBsinФ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф =  angle

Since the charged particle noves in a circle,

F = mv²/r Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the  circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ) Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C