Question

A proton is projected with a velocity of 7.0 km/s into a magnetic field of 0.60 T perpendicular to the motion of the proton. What is the magnitude of the magnetic force that acts on the proton? (e = 1.60 × 10-19 C)

Answer 1

Answer:

6.72 x 10⁻¹⁶ N

Explanation:

v = speed of the proton moving through the magnetic field = 7 km/s = 7000 m/s

B = magnitude of magnetic field in the region = 0.60 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

θ = Angle of magnetic field with the velocity of proton = 90

F = magnitude of magnetic force

magnitude of magnetic force is given as

F = q v B Sinθ

F = (1.6 x 10⁻¹⁹) (7000) (0.60) Sin90

F = 6.72 x 10⁻¹⁶ N