Question

A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 2.50 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.

Answer 1

Answer:

B = 6.18 10⁻⁶ T  

the magnetic field is in the negative direction of the y axis

Explanation:

The magnetic force is given by

         F = q v x B

as in the exercise indicate that the velocities perpendicular to the magnetic field,

         F = q v B

Newton's second law is

         F = m a

let's substitute

         q v B = m a

         B = m a / q v

let's calculate

         B = 9.1 10⁻³¹ 2.50 10¹³ / (1.6 10⁻¹⁹ 2.30 10⁷)

         B = 6.18 10⁻⁶ T

The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.

In the exercise indicate that the velocity is the z axis

the acceleration and therefore the force in the x axis

therefore the magnetic field is in the negative direction of the y axis