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# Question and Answer

## QuestionPhysicsClass 12

A cyclotron’s oscillator frequency is $$10MHz$$. What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is $$60 cm$$. what is the kinetic energy (in $$MeV$$) of the proton beam produced by the accelerator.
$$(e=1.60\times 10^{-19}C$$,$$m_p=1.67\times 10^{-27}\mathit{kg},$$  $$1\;MeV=1.6\times 10^{-31}J$$).

$$7.46\mathit{MeV}$$
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## Solution

The oscillator frequency should be same as proton’s cyclotron frequency.
According to the radius of the circular path of a particle we know that,  $$r=\frac{mv}{qB}$$  and also we know that, the larger the momentum the larger is the radius and bigger the circle described. If  $$\omega$$  is the angular frequency, then  $$\nu =\omega r$$
So,  $$\omega =2\mathit{\pi \nu }=\frac{qB}m$$
Using above equations we have,
$$B=\frac{2\pi m\upsilon }{q}$$
$$=\frac{2\times 3.14\times 1.67\times 10^{-27}\times 10^7}{1.6\times 10^{-19}}$$
$$=0.66T$$
Final velocity of proton is
$$v =r\times 2\pi \upsilon$$
$$=0.6\times 6.3\times 10^7$$
$$=3.78\times 10^7m/s$$
$$E=\frac 1 2mv^2$$
$$=\frac{1.67\times 10^{-27}\times (3.78\times 10^{7})^2}{2\times 1.6\times 10^{-19}}$$
$$=7.46\mathit{MeV}$$