A) \[2r\]
B) \[\frac{r}{2}\]
C) \[\frac{r}{\sqrt{2}}\]
D) \[r\]
Correct Answer: B
Solution :
Here: Initial radius of circular path\[{{r}_{1}}=r\] Magnetic field = B The relation for radius of circular path is given by, \[\frac{m{{\upsilon }^{2}}}{r}=Bq\upsilon \] \[r=\frac{m\upsilon }{Bq}\propto \frac{m}{q}\] As it is clear that the ratio of mass to charge (m/q) of\[\alpha -\]particle is double that of proton. Hence, \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{2}{1}\] (where\[{{r}_{2}}\]is the final radius of circular path) Or \[{{r}_{2}}=\frac{{{r}_{1}}}{2}=\frac{r}{2}\]
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