5. A proton moves perpendicularly to a uniform magnetic field, B , at 4.0 x 10 7

Question

5. A proton moves perpendicularly to a uniform magnetic field, B, at 4.0 x 107 m/s and experiences an acceleration of 1.0 x1013 m/s2 in the +x direction when its velocity is in the +z direction. Determine the magnitude and direction of the field.
Magnitude
1 TDirection +y direction +x direction -y direction -z direction -x direction +z direction 5. A proton moves perpendicularly to a uniform magnetic field, B, at 4.0 x 107 m/s and experiences an acceleration of 1.0 x1013 m/s2 in the +x direction when its velocity is in the +z direction. Determine the magnitude and direction of the field.
Magnitude
1 TDirection +y direction +x direction -y direction -z direction -x direction +z direction +y direction +x direction -y direction -z direction -x direction +z direction

Explanation / Answer

As we know

F = Bq X V

F = mass*acceleration

So

B = (1.676*10^-27*1*10^13)/(1.6*10^-19*4*10^7)

= 0.00261875 T

Direction is -Y axis

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