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?
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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
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