A protontravels through

Question

A proton travels through a region of uniform magnetic field at an angle θ relative to the magnetic field. The magnitude of the magnetic field in this region is 1.10 T, and the proton's velocity is 500,000 m/s when it experiences an acceleration whose magnitude is 1.22×10^13 m/s^2. Calculate the angle θ.

Answer 1

The formula to use is F = Q x V x B x sin θ.

Since F = ma, ma = Q x V x B x sin θ.

Look up the mass and charge of a proton, substitute into the equation along with the other given parameters, and solve for sin θ. Knowing the sine of the angle, you can calculate the angle θ.

m= 1.6726 E-27 kg, Q = 1.6 E-19 C

sin θ = 1.6726 E-27 kg x 1.22×10^13 m/s^2 / (1.6 E-19 C x 500,000 m/s x 1.10 T) = 0.232.

θ = 13.4°