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Un proton penetra perpendiculares en un campo magnetico de 5 teslas con una velocidad de 2.10 m/s calcula

Sagot :

Numenius

Answer:

The magnetic force acting on the proton is 1.68 x 10^-18 N.

Explanation:

magnetic field, B = 5 T

speed , v = 2.1 m/s

charge q = 1.6 x 10^-19 C

Angle, A = 90 degree

The magnetic force on the charge particle is given by

[tex]F = q v B sin A\\\\F = 1.6\times 10^{-19}\times 2.1\times 5\times sin 90\\\\F = 1.68\times 10^{-18} N[/tex]

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