Answer
a) $0.708 \;\rm T$
b) $5.39\;\rm MHz$
c) $273 \;\rm rev$
d) $5.06\times 10^{-5}\;\rm s$
e) $858\;\rm m$
Work Step by Step
a) We know that kinetic energy is given by
$$KE= \dfrac{q^2B^2R^2}{2m}$$
See example (32-2) in your textbook.
So, the magnetic field is given by
$$B^2=\dfrac{2mKE}{q^2R^2}$$
$$B =\sqrt{\dfrac{2mKE}{q^2R^2}}$$
Plugging the known;
$$B =\sqrt{\dfrac{2\times 2.014102\times 1.66\times 10^{-27}\times 12\times 10^6\times 1.6\times 10^{-19}}{(1.6\times 10^{-19})^2\times 1^2}}$$
$$B=\color{red}{\bf 0.708 }\;\rm T$$
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b) We know that the frequency needed is given by
$$f=\dfrac{qB}{2\pi m}$$
Plugging the known;
$$f=\dfrac{1.6\times 10^{-19}\times 0.708}{2\pi \times 2.014102\times 1.66\times 10^{-27}}$$
$$f= 5.39242\times 10^6\;\rm Hz=\color{red}{\bf5.39}\;\rm MHz$$
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c) In this case, the particles will be accelerated two times in each revolution.
Thus, the number of revolutions is given by
$$N=\dfrac{V_1}{2V_2}$$
where $V$ is the potential difference.
$$N=\dfrac{12\times 10^6}{2\times 22\times 10^3}=\color{red}{\bf 273}\;\rm rev$$
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d) We can find the time it takes by dividing the number of revolutions times the periodic time which is equal to the $1/f$.
Thus,
$$\Delta t=NT=\dfrac{N}{f}$$
Plugging the known;
$$\Delta t =\dfrac{273}{5.39\times 10^6}=\color{red}{\bf5.06\times 10^{-5}}\;\rm s$$
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e) We know that the particles are moving in circular paths through the Cyclotron and the circumferences of these paths increases in each revolution since the particles move in a larger circle and larger circle and so on..
Let's take an average circular path which has a radius equal to the rdius of the Cyclotron divided by 2.
Thereore, the distance traveled in one revolution is given by
$$d= 2\pi R_{avg}=\color{red}{\bf\not}2\pi \frac{R}{\color{red}{\bf\not}2}$$
And hence the distance traveled inside the Cyclotron is given by
$$d_{tot}=Nd=N \pi R =$$
Plugging the known;
$$d_{tot} =273\pi \times 1=\color{red}{\bf 858}\;\rm m$$
