An electron is moving at 3.0 × 10⁶ m/s perpendicular to a uniform magnetic field. If the radius of the motion is 18 mm, what is the magnitude of the magnetic field? A) 70 mT B) 0.95 mT C) 9.3 mT D) 1.6 μT E) 8.9 nT

Question

Quizplus · Accepted Answer

We can use the formula for the radius of circular motion of a charged particle in a magnetic field:

r = mv/qB

where r is the radius of the circular path, m is the mass of the particle, v is its velocity, q is its charge, and B is the magnetic field.

We can solve for B:

B = mv/qr

Plugging in the given values:

m = 9.11 x 10^-31 kg (mass of electron)
v = 3.0 x 10^7 m/s
q = -1.60 x 10^-19 C (charge of electron)
r = 0.018 m

B = (9.11 x 10^-31 kg)(3.0 x 10^7 m/s)/(-1.60 x 10^-19 C)(0.018 m)
B = 0.95 mT

Therefore, the correct answer is B.

An Electron Is Moving at 3