A proton moves in the magnetic field B = 0.46i T with a speed of 1.2 times 10^7

Question

A proton moves in the magnetic field B = 0.46i T with a speed of 1.2 times 10^7 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 F_x, F_y, F_z, 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 F_x, F_y, F_z, where the x, y, and z components are separated by commas.

Explanation / Answer

force = charge * velocity * magnetic field * sin(theta)

force = 1.6 * 10^-19 * 1.2 * 10^7 * 0.46 * sin(45)

force = 0 Fx, -6.2451671 * 10^-13 Fy, 0 Fz N

in second case

force = 1.6 * 10^-19 * 1.2 * 10^7 * 0.46 * sin(180 + 45)

force = 0 Fx, 6.2451671 * 10^-13 Fy, 0 Fz N

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