Question

.a proton enters in magnetic field of 2500N(A-m) with velocity 4.0*10 ke power 17m/s perpendicular. calculate the force acting on proton.​

Answer 1

Answer:

F= qvB

charge on protons = 1.602×10−19

B= 2500 N/A-m

velocity= 4*10^5

then F = 1.602*4*2500*10^-14

= 1.6*10^-10 N

Explanation:

I hope my answer is correctly. because I am not shiure this answer is correct...