Question #77644
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
Expert's answer
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) )
v=3.57×〖10〗^6 m/s
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) )
v=3.57×〖10〗^6 m/s
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