A proton is projected into a magnetic field of 3.00 T at a speed of 0.0667 c.?

Question

Assuming the velocity is perpendicular to the magnetic field, determine the path of the proton in the field?

Answer 1

F (Lorentz) = q (v x B) = q v B (sin theta)

In this case theta = 1 as sin 90 = 1 (so can be ignored)

q = 1.6 x 10^-19
v = 3 x 10^8 x 0.0667 = 2.001 x 10^7

F = (1.6 x 10^-19) x (2.001 x 10^7) x 3
F = 9.6048 x 10^-12 Newtons

The proton will traverse a circular path, the force governs its radius according to-:

F = m v^2 / r
r = m v^2 / F
r = (1.67 x 10^-27) x (2.001 x 10^7)^2 / (9.6048 x 10^-12)

radius = 0.06 metres or 6 cm
So it will travel in a circular path with a radius of 6cm......

Any more like this - I quite like doing Magnetic & Electric field stuff (well I do lecture it....) ;o)

Answer 2

Because F = q(v x B), we know the force will always be perpendicular to the plane of "v" and "B" (the magnetic field). If the proton gets projected into a large enough magnetic field it'll follow the path of a circle.

Answer 3

as end results of the technique of truth F = q(v x B), all individuals be conscious of the stress will continually be perpendicular to the plane of "v" and "B" (the magnetic field). If the proton receives projected suitable right into a large sufficient magnetic field it is going to persist with the route of a circle.