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
