A proton moving with a velocity of 4.0 × 10⁴ m/s enters a magnetic field of 0.20 T.If the angle between the velocity of the proton and the direction of the $\vec { B }$ field is 60°,what is the magnitude of the magnetic force exerted on the proton? (e = 1.60 × 10^-19C)

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by the equation: F = qvBsinθ where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector. In this problem, the proton has a charge of +1.60 × 10^-19C and a velocity of 4.0 × 10⁴ m/s. The magnetic field strength is given as 0.20 T, and the angle between the proton's velocity vector and the magnetic field vector is 60°. Plugging these values into the equation, we get: F = (1.60 × 10^-19C)(4.0 × 10⁴ m/s)(0.20 T)(sin 60°) F = 1.12 × 10^-15 N Therefore, the magnitude of the magnetic force exerted on the proton is 1.1 × 10^-15 N, which corresponds to choice C.

A Proton Moving with a Velocity of 4 B⃗\vec { B }B

A Proton Moving with a Velocity of 4 $\vec { B }$