A) \[B=\frac{mv}{e}\]and\[K=2m{{\pi }^{2}}{{v}^{2}}{{R}^{2}}\]
B) \[B=\frac{2\pi mv}{e}\]and\[K={{m}^{2}}\pi v{{R}^{2}}\]
C) \[B=\frac{2\pi mv}{e}\]and\[K=2m{{\pi }^{2}}{{v}^{2}}{{R}^{2}}\]
D) \[B=\frac{mv}{e}\]and\[K={{m}^{2}}\pi v{{R}^{2}}\]
Correct Answer: C
Solution :
Frequency \[v=\frac{eB}{2\pi m}\] \[KE=\frac{1}{2}m{{v}^{2}}\]and radius\[R=\frac{mv}{eB}\] Here, velocity \[v=\frac{\pi R}{T/2}=\frac{2\pi R}{T}=2\pi Rv\] \[\therefore \]Radius \[R=\frac{m(2\pi Rv)}{eB}\] Magnetic field \[B=\frac{2\pi mv}{e}\]Kinetic energy \[K=\frac{1}{2}m{{(2\pi Rv)}^{2}}=2m{{\pi }^{2}}{{v}^{2}}{{R}^{2}}\]
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