Question

A proton moving at 4.0 × 106 m / s through a magnetic field of 1.7T experiences a magnetic force of magnitude 8.2 × 10-13N What is the angle between the velocity of the proton and the field?

Answer 1

`"49 degrees"`

Explanation:

Magnetic force (`F`) is

`F = q(vecv × vecB) = BqvSintheta`

`sintheta = F/(Bqv) = (8.2 × 10^-13 "N")/("1.7 T" × "1.6 × 10"^-19 "C" × "4.0 × 10"^6 "m/s") ≈ 0.754`

`θ = "49 degrees"`