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A charged particle (Q = 6 C, m = 0.02 kg) moves at a speed of 70 m/s perpendicular to the direction of a uniform magnetic field of 0.03 T. What is the magnitude of the force exerted by the magnetic field on the charge?

Respuesta :

Answer:

Explanation:

When a charged particle moves in a magnetic field , a force acts on it perpendicular to its velocity.

Force = Bqv , B is magnetic field , q is charge of the particle and v is velocity

= .03 x 6 x 70

= 12.6 N

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