A proton enters a uniform magnetic field of strength 0.5 T at 200 m/s. The magnetic field is oriented at an angle of 30° to the proton’s velocity. What is the magnitude of the force that the proton experiences while it moves through the magnetic field?

Explanation:

When any charge moves into the magnetic field , it experiences the force

The force can be calculated by the relation F = q v B sinθ

here q is the charge and v is its velocity .

B is magnetic field strength and θ is the angle between field and velocity vectors .

Thus F = 1.6 x 10⁻¹⁹ x 200 x 0.5 x  sin 30

=  8 x 10⁻¹⁸ N

spectral lines lyman is one set

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer