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3 years ago

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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?

2 answers:

nalin [4]3 years ago

4 0

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!!!

Mrac [35]3 years ago

4 0

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

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