Question

A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of 0.00687 T 0.00687 T and the proton's velocity makes an angle of 101 ∘ 101∘ with the field, what is the magnitude of the magnetic force acting on the proton?

Answer 1

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

c = Speed of light = $3\times 10^8\ m/s$

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$