A cyclotron's oscillator frequency is $20 \mathrm{MHz}$. The operating magnetic field for accelerating…

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 $=1.67$ $\left.\mathrm{x} 10^{-27} \mathrm{~kg}\right)$
  1. $0.66 \mathrm{~T}$
  2. $1.1 \mathrm{~T}$
  3. $0.33 \mathrm{~T}$
  4. $1.31 \mathrm{~T}$

Solution

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} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

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