Question

When as proton has a velocity v=(2hati+3hatj)xx10^6m/s, it experience a force F=-(1.28xx10^-13hatk) When its velocity is along the z-axis, it experience a force along the x-axis. What is the magnetic field?

Answer 1

Thus the magnetic field is B = 0.4 j

Explanation:

Given data:

• Velocity "V" = (2i + 3j) x 10^6 m/s
• F = - (1.28 x 10^-13 k)

Solution:

F = q ( V x B)

- (1.28 x 10^-13 k) = 1.6 x 10^-19 [ (2i + 3j) x 10^6  x - Bo j]

+ 1.28 = 1.6 ( 2i + 3j x Bo j)

1.28 k = 1.6 ( 2 . Bo k)

Bo = 1.28 / 1.6 x 2 = 0.4

B = -Boj

B = 0.4 j

Thus the magnetic field is B = 0.4 j