Question In a certain cyclotron a proton moves in a circle of radius 0.390 m. The magnitude of the magnetic field is 1.50 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton? (a) Number i Units (b) Number i Units

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Transcribed Image Text: In a certain cyclotron a proton moves in a circle of radius 0.390 m. The magnitude of the magnetic field is 1.50 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton? (a) Number i Units (b) Number i Units

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a)  2.29 u00d7 107 Hz b) 2.70 u00d7 10-12 J Radius, \gamma=0.390 \mathrm{~m}magnetic fie k, B=1.50 \mathrm{~T}Mass of proton, m_{p}=1.67 \times 10^{-27} \mathrm{~kg}charge of proton, q_{p}=1.60 \times 10^{-19} \mathrm{C}(a) Centripetal force is provited by Magnetic forc. de, F_{c}=F_{B}\Rightarrow \frac{m_{p} v^{2}}{\gamma}=q_{p} v BOscillator frequency, F=\frac{V}{2 \pi r}-2\begin{array}{l}\Rightarrow \Rightarrow \frac{V}{\gamma}=\frac{q_{p} B}{m_{p}} \therefore \frac{V}{2 \pi \gamma}=\frac{q_{p} B}{2 \pi m_{p}} \\\therefore F=\frac{V}{2 \pi_{\gamma}}=\frac{q_{p} B}{2 \pi m_{p}}=\frac{1.6 \times 10^{-19}(\times 1.50 T}{2 \pi \times 1.67 \times 10^{-27} \mathrm{~kg}} \\\therefore F=2.29 \times 10 \mathrm{~Hz}\end{array}(b) speed, V=F \times 2 \pi r=2.29 \times 10^{7} H z \times 2 \pi \times 0.39 \mathrm{~m} V=5.61 \times 10^{7} \mathrm{~m} / \mathrm{s}\therefore Kinctic energy, K E=\frac{1}{2} m_{p} v^{2}=\frac{1}{2}Here, speed Is comparable to speed of light \left(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right).\therefore kinetic endrg], k=(r-1) m c^{2}where; r=\frac{1}{\sqrt{1-(v / c)^{2}}}=1 / \frac{1}{\sqrt{1-\left(\frac{5.61 \times 10 \% / 3}{3 \times 10^{3} \mathrm{~m} / \mathrm{s}}\right)^{2}}}=1.018\begin{aligned}\therefore K & =(1.018-1) \times 1.67 \times 10^{-27} \mathrm{~kg} \times\left(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2} \\& =2.70 \times 10^{-12} \mathrm{~J}\end{aligned} ...