MHT CET · Physics · Magnetic Effects of Current
An alternating electric field of frequency ' \(v^{\prime}\) is applied across the dees of a cyclotron which is used to accelerate protons of mass 'm'. The radius of the dees is 'R'. The operating magnetic field used in cyclotron is 'B'. The kinetic energy of the proton beam is given by
- A \(2 m \pi^{2} v^{2} R^{2}\)
- B \(2 m \pi v^{2} R^{2}\)
- C \(m \pi^{2} v^{2} R^{2}\)
- D \(m \pi v^{2} R^{2}\)
Answer & Solution
Correct Answer
(A) \(2 m \pi^{2} v^{2} R^{2}\)
Step-by-step Solution
Detailed explanation
Time period of cyclotron is
\(\mathrm{T}=\frac{1}{\mathrm{v}}=\frac{2 \pi \mathrm{m}}{\mathrm{eB}} ; \mathrm{B}=\frac{2 \pi \mathrm{m}}{\mathrm{e}} \mathrm{v} ; \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{eB}}=\frac{\mathrm{P}}{\mathrm{eB}}\)
\(\Rightarrow \mathrm{P}=\mathrm{eBR}=\mathrm{e} \times \frac{2 \pi \mathrm{mv}}{\mathrm{e}} \mathrm{R}=2 \pi \mathrm{mvR}\)
\(\mathrm{K.E.}\) \(=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}=\frac{(2 \pi \mathrm{mvR})^{2}}{2 \mathrm{m}}=2 \pi^{2} \mathrm{mv}^{2} \mathrm{R}^{2}\)
\(\mathrm{T}=\frac{1}{\mathrm{v}}=\frac{2 \pi \mathrm{m}}{\mathrm{eB}} ; \mathrm{B}=\frac{2 \pi \mathrm{m}}{\mathrm{e}} \mathrm{v} ; \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{eB}}=\frac{\mathrm{P}}{\mathrm{eB}}\)
\(\Rightarrow \mathrm{P}=\mathrm{eBR}=\mathrm{e} \times \frac{2 \pi \mathrm{mv}}{\mathrm{e}} \mathrm{R}=2 \pi \mathrm{mvR}\)
\(\mathrm{K.E.}\) \(=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}=\frac{(2 \pi \mathrm{mvR})^{2}}{2 \mathrm{m}}=2 \pi^{2} \mathrm{mv}^{2} \mathrm{R}^{2}\)
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