The figure is missing, so i have attached it
Answer:
Magnitude of B = 0.252 T
Step-by-step 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