Question

A charged particle moves into a region of uniform magnetic field B (pointing out of the page), goes through half a circle, and exits the region. The particle is either a proton or an electron. It spends 130 ns in the region. (a) What is the magnitude of B

Answer 1

The figure is missing, so i have attached it

Answer:

Magnitude of B = 0.252 T

Explanation:

From the image, considering the point at which it enters the field-filled region, the velocity vector is pointing downwards. The field points out of the page so that; (v→) × (B→) points leftward, points leftward which indeed seems to be the direction it is pushed. Therefore q > 0 and thus it's a proton.

The equation for the period since it goes through half circle is;

T = 2t = 2πm/(e|B|)

Where;

m is mass of proton = 1.67 × 10^(-27) kg

e is electron charge = 1.60 x 10^(-19) Coulombs.

|B| is magnitude of magnetic field

t = 130 ns = 130 × 10^(-9) s

Making |B| the subject, we have;

|B| = πm/et

Thus, plugging in all relevant values, we have;

|B| = π(1.67 × 10^(-27))/(1.60 x 10^(-19) × 130 × 10^(-9)) = 0.252 T