Homework SourseA proton moves in the magnetic field B040iT with a speed of
A proton moves in the magnetic field B?=0.40i^T with a speed of 1.0×107 m/s in the directions shown in the figure.
Part A
In Figure (a), what is the magnetic force F? on the proton? Give your answers in component form.
Express vector F? in the form of Fx , Fy , Fz , where the x , y , and z components are separated by commas.
Part B
In Figure (b), what is the magnetic force F? on the proton? Give your answers in component form.
Express vector F? in the form of Fx , Fy , Fz , where the x , y , and z components are separated by commas.
Please show all work and steps clearly, thank you.
Solution
Part A
F=qvBsin (theta)
=(1.602*10^-19)(1*10^7 m/s k^)(0.40T i^)sin45
=0.458*10^-12 N j^
b)
F=qvBsin (theta)
=(1.602*10^-19)(1*10^7 m/s -j^)(0.40Ti^)sin270
= 0.648 N k^ (in positive z direction.)
