A proton has a speed of 3.0 × 10⁶ m/s in a direction perpendicular to a uniform magnetic field, and the proton moves in a circle of radius 0.20 m. What is the magnitude of the magnetic field? A) 0.080 T B) 0.16 T C) 0.24 T D) 0.32 T E) 0.36 T

Question

Quizplus · Accepted Answer

The equation for the radius of a circular path for a charged particle in a magnetic field is:

r = mv / (qB)

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

Solving for B, we get:

B = mv / (qr)

Plugging in the values given in the problem, we get:

B = (1.67 × 10^-27 kg) (3.0 × 10^7 m/s) / [(1.60 × 10^-19 C) (0.20 m)]

B = 0.16 T

Therefore, the answer is B) 0.16 T.

A Proton Has a Speed of 3