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A proton moving at 2.5 × 10^3 m/s horizontally enters a region where a magnetic field of 0.42 T is present, directed vertically upward. What magnitude and direction of force acting on the proton due to this field? (e= 1.60 × 10^-19 C)

Respuesta :

Karabo99

Answer:

F  = 1.68×10^-16 N, in the - y direction.

Explanation:

let v be the speed of the proton, q be the charge of the proton and B be the magnetic field.

Then, the force acting on the proton is given by:

F = q×B×v  , F,B and v are vectors.

  = (1.60×10^-19)×(0.42)×(2.5×10^3)

  = 1.68×10^-16 N

since: F is the cross product of B and v by the right hand rule, choosing the direction of the speed of the proton to be in the x-axis and the direction of the magnetic field to be in the z-axis then the force will be the - y axis.

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