A) \[\frac{1}{\sqrt{2}}\]
B) \[\sqrt{2}\]
C) \[\frac{3}{\sqrt{2}}\]
D) \[\frac{2}{\sqrt{3}}\]
Correct Answer: A
Solution :
The radius of circular path is given by \[r=\frac{\mu \upsilon }{qB}\] If \[{{E}_{K}}\]be kinetic energy of particle then its momentum \[p=m\upsilon =\sqrt{2m{{E}_{K}}}\] So, the radius is given by \[r=\sqrt{\frac{2m{{E}_{K}}}{qB}}\] So, \[r\propto \sqrt{{{E}_{K}}}\] Therefore, when kinetic energy is halved the radius is reduced to \[\frac{1}{\sqrt{2}}\] times its initial value.
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