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A cyclotron's oscillator frequency is \( 10 \mathrm{MHz} \) and the operating magnetic field is \( 0.66 \mathrm{~T} \). If the
radius of its dees is \( 60 \mathrm{~cm} \), then the kinetic energy of the proton beam produced by the
accelerator is
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radius of its dees is \( 60 \mathrm{~cm} \), then the kinetic energy of the proton beam produced by the
accelerator is
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Verified Answer
The correct answer is:
\( 7 \mathrm{Mev} \)
We know, radius of circular path
$r=\frac{m v}{q B}$
$\Rightarrow v=\frac{q B r}{m} \rightarrow(1)$
Frequency of cyclotron, $f=\frac{q B}{2 \Pi m}$
$\Rightarrow B=\frac{2 \Pi m f}{q} \rightarrow(2)$
Using Eq. (1), kinetic energy $=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{q^{2} B^{2} r^{2}}{m^{2}}$
Using Eq. (2), kinetic energy $=\frac{1}{2} \frac{q^{2} r^{2}}{m} \frac{(2 \Pi)^{2} m^{2} f^{2}}{q^{2}}=2 \Pi^{2} f^{2} r^{2} m$
Therefore, kinetic energy $=2 \times(3.14)^{2} \times\left(10 \times 10^{6}\right)^{2} \times\left(60 \times 10^{-2}\right)^{2} \times\left(1.67 \times 10^{-27}\right)$
$=1.13 \times 10^{-12} \mathrm{~J}$
Hence, kinetic energy
$=\frac{1.13 \times 10^{-12}}{1.16 \times 10^{-19} \times 10^{6}} \simeq 7.1 \mathrm{MeV} \approx 7 \mathrm{MeV}$
$r=\frac{m v}{q B}$
$\Rightarrow v=\frac{q B r}{m} \rightarrow(1)$
Frequency of cyclotron, $f=\frac{q B}{2 \Pi m}$
$\Rightarrow B=\frac{2 \Pi m f}{q} \rightarrow(2)$
Using Eq. (1), kinetic energy $=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{q^{2} B^{2} r^{2}}{m^{2}}$
Using Eq. (2), kinetic energy $=\frac{1}{2} \frac{q^{2} r^{2}}{m} \frac{(2 \Pi)^{2} m^{2} f^{2}}{q^{2}}=2 \Pi^{2} f^{2} r^{2} m$
Therefore, kinetic energy $=2 \times(3.14)^{2} \times\left(10 \times 10^{6}\right)^{2} \times\left(60 \times 10^{-2}\right)^{2} \times\left(1.67 \times 10^{-27}\right)$
$=1.13 \times 10^{-12} \mathrm{~J}$
Hence, kinetic energy
$=\frac{1.13 \times 10^{-12}}{1.16 \times 10^{-19} \times 10^{6}} \simeq 7.1 \mathrm{MeV} \approx 7 \mathrm{MeV}$
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