A proton having a speed of 3.0 × 10⁶ m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field.What is the magnitude of the $\vec { B }$ field? (e = 1.60 × 10^-19 C,m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

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

r = mv / (|q|B)

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

Solving for B, we get:

B = mv / (|q|r)

Substituting the given values, we get:

B = (1.67 × 10^-27 kg) × (3.0 × 10^7 m/s) / (|1.60 × 10^-19 C| × 0.20 m) = 0.16 T

Therefore, the correct answer is B) 0.16 T.

A Proton Having a Speed of 3 B⃗\vec { B }B

A Proton Having a Speed of 3 $\vec { B }$