Respuesta :
Answer:
F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N
Magnitude of F = (2.466 × 10⁻¹⁸) N
Explanation:
The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product
F = qv × B
where v = (4î − 6ĵ + k) m/s
B = (î + 2ĵ − k) T
The particle is a proton, hence,
q = (1.602 × 10⁻¹⁹) C
F = qv × B = q (v × B)
(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)
The cross product is evaluated as a determinant of
| î ĵ k |
|4 -6 1 |
|1 2 -1 |
î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]
î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)
(v × B) = (4î + 5ĵ + 14k)
F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)
F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N
Magnitude of F =
√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]
Magnitude of F = (2.466 × 10⁻¹⁸) N
Hope this Helps!!!
Answer:
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
