A) \[\frac{\sqrt{2m\left( E-{{W}_{0}} \right)}}{eB}\]
B) \[\sqrt{2m\left( E-{{W}_{0}} \right)}eB\]
C) \[\frac{\sqrt{2e\left( E-{{W}_{0}} \right)}}{mB}\]
D) \[\frac{\sqrt{2m\left( E-{{W}_{0}} \right)}}{eB}\]
Correct Answer: D
Solution :
From Einsteins equation \[E={{W}_{0}}+\frac{1}{2}m{{v}^{2}}\] \[\sqrt{\frac{2(E-{{W}_{0}})}{m}}=v\] or A charged particle placed in uniform magnetic field experience a force \[F=\frac{m{{v}^{2}}}{r}\] or \[evB=\frac{m{{v}^{2}}}{r}\] or \[r=\frac{mv}{eB}\] or \[r=\frac{m\sqrt{\frac{2(E-{{W}_{0}})}{m}}}{eB}\] \[\Rightarrow \] \[r=\frac{\sqrt{2m(E-{{W}_{0}})}}{eB}\]You need to login to perform this action.
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