A proton moves with a velocity of = (3î − 4ĵ + ) m/s in a region in which the magnetic field is = (î + 2ĵ − ) T. What is the magnitude of the magnetic force this particle experiences?
Question
A proton moves with a velocity of = (3î − 4ĵ + ) m/s in a region in which the magnetic field is = (î + 2ĵ − ) T. What is the magnitude of the magnetic force this particle experiences?
Solution
The magnetic force experienced by a moving charged particle in a magnetic field is given by the equation:
F = q(v × B)
where: F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.
The proton has a charge of q = +1.6 x 10^-19 C.
The velocity v = 3î - 4ĵ + 0k (m/s) and the magnetic field B = 1î + 2ĵ - 0k (T).
The cross product of v and B is:
v × B = (3î - 4ĵ) × (1î + 2ĵ) = 3î × 1î + 3î × 2ĵ - 4ĵ × 1î - 4ĵ × 2ĵ = 0k + 6k - 4k + 0k = 2k
So, the magnetic force F = q(v × B) = (1.6 x 10^-19 C)(2 T.m/s) = 3.2 x 10^-19 N.
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