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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)
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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)
Solution:
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Verified Answer
The correct answer is:
$\frac{2 \pi n m}{q}, \frac{q v B r}{2}$
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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