Question

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.

Answer 1

B = 0.013(-j) T

Step-by-step explanation:

Given that,

The speed of a proton,
`v=1.75* 10^7\ m/s`

Acceleration experienced by the proton,
`a=2.25* 10^3\ m/s`

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

`ma=qvB\\\\B=(ma)/(qv)\\\\B=(1.67* 10^(-27)* 2.25* 10^(13))/(1.6* 10^(-19)* 1.75* 10^(7))\\\\B=0.013\ T`

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