A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 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.

B = 0.013(-j) T

Explanation:

Given that,

The speed of a proton,

Acceleration experienced by the proton,

We need to find the magnitude and the direction of the magnetic field. At equilibrium,

The velocity is in +z direction, force in +x direction, then the field must be in -y direction.

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