Question

An electron enters a region of space containing a uniform 2.71 × 10 − 5 2.71×10−5 T magnetic field. Its speed is 197 197 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion.

Answer 1

Answer:

r = 0.0414mm

F = 757,692.3Hertz

Explanation:

If the body enters space with uniform magnetic field B, the force experienced by the object is expressed as

F = qvBsintheta... 1

Also, if the body undergoes a circular motion, the force experienced by the body in a circular path is given as

Fc = mv²/r... 2

Equating both forces

F = Fc

qvBsin theta = mv²/r

Since the body enters perpendicular to the field, theta = 90°

The equality becomes;

qvB sin90° = mv²/r

qvB = mv²/r

qB = mv/r

r = mv/qB

Given mass of the electron m = 9.11×10^-31kg

Velocity of the object v = 197m/s

Charge on the electron q = 1.6×10^-19C

Magnetic field B = 2.71×10^-5T

Substituting this value into the equation to get the radius r we have;

r = 9.11×10^-31 × 197/1.6×10^-19 × 2.71×10^-5

r = 1794.67×19^-31/4.336×10^-24

r = 413.89×10^-7

r = 0.0000414m

r = 0.0414mm

b) Frequency of the motion F = w/2π where w is the angular velocity

Since w = v/r

F = (v/r)/2π

F = v/2πr

F = 197/2π(0.0000414)

F = 757,692.3Hertz