A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î + 2ĵ − k) T. What is the magnitude of the magnetic force this particle experiences?

1.442×10^-18N

Explanation:

Magnitude of magnetic force experienced by the particle id expressed as;

F = qvB sin theta where;

q is the charge on the proton = 1.602×10^-19C

v is the velocity =(4î − 6ĵ + k) m/s

B is the magnetic field = (î + 2ĵ − k) Tesla

theta is the angle that the conductor make with the magnetic field = 90°

Substituting this values in the formula to get the magnitude of the force we have;

F = 1.602×10^-19 × (4î − 6ĵ + k)(î + 2ĵ − k) sin90°

Multiplying the vector function

(4î − 6ĵ + k)(î + 2ĵ − k) = 4i.i - 12j.j - k.k

Since i.i = j.j = k.k = 1

(4î − 6ĵ + k)(î + 2ĵ − k) = 4-12-1 = -9

F = 1.602×10^-19×(-9)×1

F = -14.418×10^-19N

F = -1.442 × 10^-18N

|F| = 1.442 × 10^-18N

The magnitude of the magnetic force this particle experiences is therefore 1.442 × 10^-18N

report!

explanation:

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