The magnetic poles of a small cyclotron produce a magnetic field with magnitude \(0.85\,{\rm{T}}\). The poles have a radius of \(0.40\,{\rm{m}}\), which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons (\(q = 1.60 \times {10^{ - 19}}\,{\rm{C}}\), \(m = 1.67 \times {10^{ - 27}}\,{\rm{kg}}\)) can be accelerated by this cyclotron? Give your answer in electron volts and in joules. (b) What is the time for one revolution of a proton orbiting at this maximum radius? (c) What would the magnetic-field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part (a)? (d) For \(B = 0.85\,{\rm{T}}\), what is the maximum energy to which alpha particles (\(q = 3.20 \times {10^{ - 27}}\,{\rm{C}}\), \(m = 6.64 \times {10^{ - 27}}\,{\rm{kg}}\)) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

The force acting on the cyclotron is given by

\({F_{\rm{B}}} = m{a_{{\rm{rad}}}}\)

And

\(|q|vB = \frac{{m{v^2}}}{R}\)

The radius of the pole is, \({R_{\max }} = 0.40\,{\rm{m}}\).

The value of magnetic field is, \(B = 0.85\,{\rm{T}}\).

(a)

The force acting on the cyclotron is given by

\({F_B} = m{a_{rad}}\)

Therefore,

\(|q|vB = \frac{{m{v^2}}}{R}\)

Now the maximum velocity is,

\({v_{\max }} = \frac{{|q|B{R_{\max }}}}{m}\)

Substitute all the value in the above equation.

\(\begin{aligned}{v_{\max }} = \frac{{\left( {1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)(0.85\,{\rm{T}})(0.40\,{\rm{m}})}}{{1.67 \times {{10}^{ - 27}}\,{\rm{kg}}}}\\ = 3.257 \times {10^7}\,{\rm{m/s}}\end{aligned}\)

So, the maximum energy will be

\({K_{\max }} = \frac{1}{2}m{v_{\max }}\)

Substitute all the value in the above equation.

\(\begin{aligned}{K_{\max }} = \frac{1}{2}\left( {1.67 \times {{10}^{ - 27}}\,{\rm{kg}}} \right){\left( {3.257 \times {{10}^7}\,{\rm{m/s}}} \right)^2}\\ = 8.9 \times {10^{ - 13}}\,{\rm{J}}\\{K_{\max }} = 5.5\,{\rm{MeV}}\end{aligned}\)

Therefore, the maximum energy is \({K_{\max }} = 5.5\,{\rm{MeV}}\)

(b)

The period is one-time revolution is

\(T = \frac{{2\pi R}}{v}\)

The maximum radius is \({R_{\max }}\)

The time period is,

\({T_{\max }} = \frac{{2\pi {R_{\max }}}}{v}\)

Substitute all the value in the above equation.

\(\begin{aligned}{T_{\max }} = \frac{{2\pi (0.40\,{\rm{m}})}}{{3.257 \times {{10}^7}{\rm{m/s}}}}\\{T_{\max }} = 7.7 \times {10^{ - 8}}{\rm{s}}\end{aligned}\)

Therefore, the time for one revolution of a proton orbiting at this maximum radius is\(7.7 \times {10^{ - 8}}{\rm{s}}\)

(c)

The velocity is

\(v = \frac{{|q|BR}}{m}\)

The kinetic energy is

\(\begin{aligned}K = \frac{1}{2}m{v^2}\\ = \frac{1}{2}m{\left( {\frac{{|q|BR}}{m}} \right)^2}\\ = \frac{1}{2}\frac{{|q{|^2}{B^2}{R^2}}}{m}\end{aligned}\)

Or,

\(B = \frac{{\sqrt {2Km} }}{{|q|R}}\)

Substitute all the value in the above equation.

Therefore, the magnitude is

\(\begin{aligned}B = \sqrt 2 (0.85\,{\rm{T}})\\ = 1.2\,{\rm{T}}\end{aligned}\)

Therefore, the magnitude of the magnetic field is \(1.2\,{\rm{T}}\)

(d)

The maximum velocity is,

\({v_{\max }} = \frac{{|q|BR}}{m}\)

Substitute all the values in the above equation.

\(\begin{aligned}{v_{\max }} = \frac{{\left( {3.20 \times {{10}^{ - 19}}\,{\rm{C}}} \right)(0.85\,{\rm{T}})(0.40\,{\rm{m}})}}{{6.65 \times {{10}^{ - 27}}\,{\rm{kg}}}}\\ = 1.636 \times {10^7}\,{\rm{m/s}}\end{aligned}\)

The maximum kinetic energy is,

\({K_{\max }} = \frac{1}{2}m{v_{{{\max }^2}}}\)

Substitute all the value in the above equation.

\(\begin{aligned}{K_{\max }} = \frac{1}{2}\left( {6.65 \times {{10}^{ - 27}}\,{\rm{kg}}} \right){\left( {1.636 \times {{10}^7}\,{\rm{m/s}}} \right)^2}\\ = 8.9 \times {10^{ - 13}}\,{\rm{J}}\\ = 5.5\,{\rm{MeV}}\end{aligned}\)

Therefore, the maximum kinetic energy is \(5.5\,{\rm{MeV}}\)