A proton moves perpendicularly to a magnetic field that has a magnitude of 4.20*10^-2=T What is the speed of the particle if magnitude of magnetic force on it is 2.40*10^-14N
1
Expert's answer
2018-05-29T10:53:08-0400
Use the expression for the Lorentz force F_L=qvBsinα α=90° sin90°=1 So, we get F_L=qvB Find the v v=F_L/qB v=(2.4×〖10〗^(-14) T)/((1.6×〖10〗^(-19) C)× (4.2×〖10〗^(-2) N) )
Comments
Leave a comment