If the cyclotron oscillator frequency is 16 MHz , then what should be the operating magnetic field for accelerating the proton of mass 1 .67 × 10 - 27 kg ?

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TestHub · Accepted Answer

Given,

Frequency of cyclotron oscillator $= 16 MHz$

$v = 16 \times 10^{6} Hz$

Mass of proton, $m_{p} = 1.67 \times 10^{- 27}$

The cyclotron frequency,

$v = \frac{B q}{2 π m}$

Where $B =$ Required magnetic field

$B = \frac{v \times 2 π m}{q}$

$= \frac{16 \times 10^{6} \times 2 \times π \times 1 .67 \times 10^{- 27}}{1 .6 \times 10^{- 19}} = 0 .334 π T$

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