A charged particle having electric charge Q = 4.0e enters a region of magnetic field 1.2 T at an angle of θ relative to the magnetic field direction. If the speed of the particle upon entering the region is 2.8 × 10⁴ m/s, and the magnitude of the force on it is 1.3 × 10^-14 N, which of the following could be the angle θ? A) 47° B) 37° C) 53° D) 65°

Question

Quizplus · Accepted Answer

The force on a charged particle moving through a magnetic field is given by the equation F = qvBsin(θ), where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.

Solving for θ gives sin(θ) = F/(qBv). Plugging in the given values yields sin(θ) = (1.3 × 10^-14 N)/(4.0e * 1.2 T * 2.8 × 10^7 m/s) ≈ 0.137.

Taking the inverse sine of this value gives θ ≈ 7.9°. Since the given angles are all greater than this value, we can eliminate choices A, C, and D.

The only remaining choice is B, which gives sin(37°) ≈ 0.601. Plugging this into the equation for the force gives F = (4.0e * 1.2 T * 2.8 × 10^7 m/s) * 0.601 ≈ 1.3 × 10^-14 N, which matches the given force value. Therefore, the answer is B.

A Charged Particle Having Electric Charge Q = 4