Aproton moving at 8.9 × 106 m/s through a magnetic field of 0.96 t experiences a magnetic force of magnitude 3.8 × 10−13 n. what is the angle between the proton’s velocity and the field? the charge on a proton is 1.60218 × 10−19 c and its mass is 1.67262 × 10−27 kg.

15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then  α =15.66°

need to see circuit to get time constant

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000