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Magnetic Field and Direction ?
Part 1 of 2:
A proton moves perpendicularly to a uni- form magnetic field B with a speed of 2 × 107 m/s and experiences an acceleration of 2.6 × 1013 m/s2 in the positive x direction when its velocity is in the positive z direction.The mass of a proton is 1.673 × 10−27 kg. Find the magnitude of the field.Part 2 of 2: What is its direction?Positive or negative and in the Z, Y, or X direction
1 Answer
- ?Lv 74 weeks agoFavorite Answer
1) force = m*a = q*v*B → → B = ma / qv
B = 1.673E-27kg * 2.6E13m/s² / (1.6E-19C * 2E7m/s)
B = 1.36E-2 T = 13.6 mT
2) directionally, v x B = a → where x is the cross product
For k^ x B = i^, it must be that B = - j^ --
the field points in the -Y direction.
