A) \[0.9\times {{10}^{6}}m\text{/}s\]
B) \[2.1\times {{10}^{4}}m\text{/}s\]
C) \[1.5\times {{10}^{3}}m\text{/}s\]
D) \[1.9\times {{10}^{5}}m\text{/}s\]
Correct Answer: C
Solution :
As, \[{{m}_{e}}=9\times {{10}^{-31}}kg,\]\[{{r}_{e}}=0.02\,m\] and \[{{v}_{e}}=3\times {{10}^{6}}m\text{/}s\] Also \[{{m}_{c}}=\]mass of charged particle \[=1.8\times {{10}^{-27}}kg\] \[{{r}_{c}}=0.02\,m,\,\,{{v}_{c}}=?\] For the motion of charge in the field, \[qvB=\frac{m{{v}^{2}}}{r}\] \[v=\frac{qrB}{m}\Rightarrow v\propto \frac{1}{m}\] Thus, we have \[\frac{{{v}_{c}}}{{{v}_{e}}}=\frac{{{m}_{e}}}{{{m}_{c}}}\] \[\Rightarrow \] \[{{v}_{c}}=\frac{{{m}_{e}}}{{{m}_{c}}}\times {{v}_{e}}=\frac{9\times {{10}^{-31}}\times 3\times {{10}^{6}}}{1.8\times {{10}^{-27}}}\] \[\Rightarrow \] \[{{v}_{c}}=1.5\times {{10}^{3}}m\text{/}s\]
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