Answered

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In a certain cyclotron a proton moves in a circle of radius 0.530 m. The magnitude of the magnetic field is 1.30 T. (a) What is the oscillator frequency

Sagot :

Answer:

[tex]f=1.98\times 10^7\ Hz[/tex]

Explanation:

Given that,

The radius of circle, r = 0.53 m

The magnitude of the magnetic field, B = 1.3 T

We need to find the oscillator frequency. It is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

Put all the values,

[tex]f=\dfrac{1.6\times 10^{-19}\times 1.3}{2\pi \times 1.67\times 10^{-27}}\\\\f=1.98\times 10^7\ Hz[/tex]

So, the oscillator frequency is [tex]1.98\times 10^7\ Hz[/tex].

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