A) 0.053 m
B) 0.136 m
C) 0.157m
D) 0.236m
Correct Answer: C
Solution :
Given, \[B=8.35\times {{10}^{-2}}iT\] \[v=(2\times {{10}^{5}}i+4\times {{10}^{5}}j)\] The distance covered by proton \[P=T\,(v)\] \[=2\pi \frac{m}{qB}(v)\] \[=2\times 3.14\times \frac{1.67\times {{10}^{-27}}}{1.6\times {{10}^{-19}}\times 8.35\times {{10}^{-2}}i}\] \[\times (2\times {{10}^{5}}i+4\times {{10}^{5}}j)\] P = 0.157 m (Mass of proton \[=1.67\times {{10}^{-27}}kg\])You need to login to perform this action.
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