A proton moves in the magnetic field B ?=0.40 i ^T with a speed of 1.0×10 7 m/s

Question

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.

Explanation / Answer

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.)

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