A proton moves perpendicularly to a uniform magneticfield (Vector B) at 1.0 x10

Question

A proton moves perpendicularly to a uniform magneticfield (Vector B) at 1.0 x107 m/s and exhibits an acceleration of 2.0 x 1013 m/s2 in the+x-direction when its velocity is in the+z-direction. Determine the magnitude and direction of thefield.
1 T
2---Select----y direction+z direction-x direction+x direction+y direction-z direction ? -y direction +z direction -x direction +x direction +y direction -z direction A proton moves perpendicularly to a uniform magneticfield (Vector B) at 1.0 x107 m/s and exhibits an acceleration of 2.0 x 1013 m/s2 in the+x-direction when its velocity is in the+z-direction. Determine the magnitude and direction of thefield.
1 T
2---Select----y direction+z direction-x direction+x direction+y direction-z direction ? -y direction +z direction -x direction +x direction +y direction -z direction

Explanation / Answer

charge q = 1.6 * 10 ^ -19 C speed v = 1 * 10 ^ 7 m / s accleration a = 2 * 10 ^ 13 m / s ^ 2 force on proton   F = ma where m = mass of proton = 1.67 * 10 ^ -27 kg So, F = 3.3 * 10 ^ -14 N we know F = Bvq from this magnetic field B = F / vq                                        = 2.0875 * 10 ^ -2 T                                        = 0.20875 * 10 ^ -3 T                         

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