A proton moves in the magnetic field B =0.44 i ^Twith a speed of 1.2×10 7 m/s in

Question

A proton moves in the magnetic field B =0.44i^Twith a speed of 1.2×107 m/s in the directions shown in the figure. (Figure 1)

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.

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.

Explanation / Answer

F= V x B (vector product)

F = (q)(V)(B)sin(theta)

a) F = (1.6x10-19)(1.2x107)(.44)(.707)

F = 0.598 x 10-12 j^ N

b) F = (1.6x10-19)(1.2x107)(.44)(1)

F = 0.846 x 10-12 k^ N

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