A magnetic field forces an electron to move in a circle withradial acceleration 3.0 × 10 14 m/s2. (a) What is the speed of theelectron if the radius of its circular path is 15 cm? (b) What is theperiod of the motion?
Question
A magnetic field forces an electron to move in a circle withradial acceleration 3.0 × 10 14 m/s2. (a) What is the speed of theelectron if the radius of its circular path is 15 cm? (b) What is theperiod of the motion?
Solution
Sure, let's solve this step by step.
(a) We know that the radial or centripetal acceleration a of an object moving in a circle of radius r at a speed v is given by a = v²/r. We can rearrange this to solve for v: v = sqrt(ar).
Substituting the given values:
v = sqrt((3.0 × 10^14 m/s²)(0.15 m)) v = sqrt(4.5 × 10^13 m²/s²) v = 6.7 × 10^6 m/s
So, the speed of the electron is 6.7 × 10^6 m/s.
(b) The period T of the motion is the time it takes for the electron to complete one full circle. We can find this using the formula T = 2πr/v.
Substituting the given values:
T = 2π(0.15 m) / (6.7 × 10^6 m/s) T = 0.14 × 10^-6 s
So, the period of the motion is 0.14 microseconds.
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