Una partícula alfa (la cual es un núcleo del átomo de helio) se mueve hacia el norte con una velocidad de 3.8 X 10^5 m/s en una región donde el campo magnético es 1.9 T y está apuntando horizontalmente hacia el este. ¿Cuál es la magnitud y la dirección de la fuerza magnética sobre esta partícula alfa?

 F = 23.1 10⁻¹⁴ N,  points inside the page

Explanation:

The magnetic force is given by the expression

          F = q v x B

bold letters indicate vectors, the magnitude of this formula is

         F = q v B  sin θ

in this case the magnetic field is in the east direction and the particle moves north, so the angle is 90 and the sine 90 = 1

         F = q v B

         F = 2 1.6 10⁻¹⁹  3.8 10⁵ 1.9

         F = 23.1 10⁻¹⁴ N

the direction of the force is given by the right hand rule, for a positive charge

the thumb points in the directional of the speed, in this case towards the North

the other fingers extended in the direction of the magnetic field, in this case to the East

the palm points in the direction of the force, consequently it points inside the page,

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.