The oscillator frequency of a cyclotron is 10 | vedclass

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(D) The cyclotron frequency $f$ is given by the formula: $f = \frac{qB}{2 \pi m}$.Rearranging the formula to solve for the magnetic field $B$,we get:

Vedclass · Accepted Answer

$0.656$

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(D) The cyclotron frequency $f$ is given by the formula: $f = \frac{qB}{2 \pi m}$.Rearranging the formula to solve for the magnetic field $B$,we get: $B = \frac{2 \pi m f}{q}$.Given values are: $f = 10 \,MHz = 10^7 \,Hz$,$m = 1.67 	imes 10^{-27} \,kg$,and $q = 1.6 	imes 10^{-19} \,C$.Substituting these values into the equation:$B = \frac{2 	imes 3.14159 	imes 1.67 	imes 10^{-27} 	imes 10^7}{1.6 	imes 10^{-19}}$.$B = \frac{10.493 	imes 10^{-20}}{1.6 	imes 10^{-19}}$.$B \approx 0.656 \,T$.Thus,the correct option is $D$.

The oscillator frequency of a cyclotron is $10 \,MHz$. What should be the operating magnetic field for accelerating protons (in $,T$)? (Given: mass of proton $m = 1.67 \times 10^{-27} \,kg$,charge of proton $q = 1.6 \times 10^{-19} \,C$)

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