Question
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, calculate the kinetic energy (in MeV) of the proton beam produced by the accelerator.
✓
Answer
Magnetic field $\text{B} = \frac{2\pi\text{mf}}{\text{q}}$
$ = \frac{2\times3.14 \times1.67\times 10^{−27}\times 10^{7}}{1.6\times 10^{−19}} = 0.66\text{T}$
Final velocity of proton $\text{v} =\text{R} \times2\pi\text{f} = 0.6\times 2 \times 3.14 \times 10^{7} = 3.77 \times 10^{7}\text{𝑚/𝑠}$
Energy $ =\frac{1}{2}\text{m}v^{2}=\frac{1}{2}\times 1.67 \times 10^{−27}\times (3.77 \times 10^{7})^2\text{𝑗} = 7.4\text{ 𝑀𝑒𝑉}.$
$ = \frac{2\times3.14 \times1.67\times 10^{−27}\times 10^{7}}{1.6\times 10^{−19}} = 0.66\text{T}$
Final velocity of proton $\text{v} =\text{R} \times2\pi\text{f} = 0.6\times 2 \times 3.14 \times 10^{7} = 3.77 \times 10^{7}\text{𝑚/𝑠}$
Energy $ =\frac{1}{2}\text{m}v^{2}=\frac{1}{2}\times 1.67 \times 10^{−27}\times (3.77 \times 10^{7})^2\text{𝑗} = 7.4\text{ 𝑀𝑒𝑉}.$
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