A) 0.053 m
B) 0.136 m
C) 0.157 m
D) 0.236 m
Correct Answer: C
Solution :
(c) 0.157 m Here, \[\overrightarrow{B}=8.35\times {{10}^{-2}}\widehat{i}\,T\] \[\overrightarrow{v}=2\times {{10}^{5}}\widehat{i}+4\times {{10}^{5}}\widehat{j}\,m/s,\,m=1.67\times {{10}^{-27}}kg\] Pitch of the helix (i.e., the linear distance moved along the magnetic field in one rotation) is given by Pitch of the helix \[=\frac{2\pi mvx}{qB}\] \[=\frac{2\times 3.14\times 1.67\times {{10}^{-27}}\times 2\times {{10}^{5}}}{1.6\times {{10}^{-19}}\times 8.35\times {{10}^{-2}}}=0.157\,m\]You need to login to perform this action.
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