MHT CET · Physics · Magnetic Effects of Current
A cyclotron's oscillator frequency is 'n' and radius of the dees is 'r'. The operating
magnetic field (B) for accelerating protons of charge ' \(\mathrm{q}^{\prime}\) and kinetic energy of
protons produced by the accelerator is respectively ('m' and ' \(\mathrm{v}\) ' be the mass and
velocity of proton)
- A \(\frac{2 \pi n m}{q}, \frac{q v B r}{2}\)
- B \(\frac{\pi \mathrm{nm}}{\mathrm{q}}, \frac{\mathrm{qvBr}}{2}\)
- C \(\frac{2 \pi n m}{q}, q v B r\)
- D \(\frac{4 \pi \mathrm{nm}}{\mathrm{q}}, \frac{\mathrm{qvBr}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 \pi n m}{q}, \frac{q v B r}{2}\)
Step-by-step Solution
Detailed explanation
Cyclotron frequency \(\mathrm{n}=\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
\(\therefore \mathrm{B}=\frac{2 \pi \mathrm{nm}}{\mathrm{q}}\)
Also, \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{q} \mathrm{vB} \quad \therefore \mathrm{mv}^{2}=\mathrm{q} \mathrm{vB} \mathrm{r}\) K.E. \(=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{qvBr}}{2}\)
\(\mathrm{K} . \mathrm{E} .=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{qvBr}}{2}\)
\(\therefore \mathrm{B}=\frac{2 \pi \mathrm{nm}}{\mathrm{q}}\)
Also, \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{q} \mathrm{vB} \quad \therefore \mathrm{mv}^{2}=\mathrm{q} \mathrm{vB} \mathrm{r}\) K.E. \(=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{qvBr}}{2}\)
\(\mathrm{K} . \mathrm{E} .=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{qvBr}}{2}\)
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