A) \[R\]
B) \[R\sqrt{2}\]
C) \[R/\sqrt{2}\]
D) \[2R\]
Correct Answer: B
Solution :
Energy of charged particle revolving in a magnetic field, \[E=\frac{1}{2}m{{\upsilon }^{2}}\] But \[\frac{m{{\upsilon }^{2}}}{R}=q\upsilon B\] \[\Rightarrow \] \[\upsilon =\frac{qBR}{m}\] \[\therefore \] \[E=\frac{1}{2}m\frac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{{{m}^{2}}}\] \[=\frac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2m}\] \[\Rightarrow \] \[E\propto {{R}^{2}}\] \[\Rightarrow \] \[R\propto \sqrt{E}\] If R' is new radius \[\frac{R'}{R}=\sqrt{\frac{E'}{E}}\] \[\Rightarrow \] \[\frac{R'}{R}=\sqrt{2}\] \[\Rightarrow \] \[R'=\sqrt{2}R\]
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