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A cyclotron's oscillator frequency is $20 \mathrm{MHz}$. The operating magnetic field for accelerating protons is (charge of proton $=1.6 \times 10^{-19} \mathrm{C}$, mass of proton $=1.67$ $\left.\mathrm{x} 10^{-27} \mathrm{~kg}\right)$
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Verified Answer
The correct answer is:
$1.31 \mathrm{~T}$
We have
$$
\begin{aligned}
& \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{mR} \omega}{\mathrm{qB}} \Rightarrow \mathrm{B}=\frac{\mathrm{m} \times 2 \pi \mathrm{f}}{\mathrm{q}} \\
& =\frac{1.67 \times 10^{-27} \times 2 \times 3.14 \times 20 \times 10^6}{1.6 \times 10^{-19}}=1.31 \mathrm{~T}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{mR} \omega}{\mathrm{qB}} \Rightarrow \mathrm{B}=\frac{\mathrm{m} \times 2 \pi \mathrm{f}}{\mathrm{q}} \\
& =\frac{1.67 \times 10^{-27} \times 2 \times 3.14 \times 20 \times 10^6}{1.6 \times 10^{-19}}=1.31 \mathrm{~T}
\end{aligned}
$$
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