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Question 31 of 46

Q.A cyclotron is used to accelerate protons to a kinetic energy of 5 MeV. If the strength of magnetic field in the cyclotron is 2 T, find the radius and the frequency needed for the applied alternating voltage of the cyclotron. (Given : Velocity of proton =3×107= 3\times10^7 m/s)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019Subjective· 3mImportance★★★★★
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In a cyclotron, the magnetic force provides the centripetal force for circular motion: r=mv/(qB)r=mv/(qB) and cyclotron frequency f=qB/(2πm)f=qB/(2\pi m).

In a cyclotron, a charged particle moves in a circle because the magnetic force provides the centripetal force:

qvB=mv2r⇒r=mvqBqvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB}

Given: proton mass m=1.67×10−27 kgm = 1.67\times10^{-27}\ \text{kg}, charge q=1.6×10−19 Cq=1.6\times10^{-19}\ \text{C}, v=3×107 m/sv = 3\times10^{7}\ \text{m/s}, B=2 TB = 2\ \text{T}.

r=1.67×10−27×3×1071.6×10−19×2=5.01×10−203.2×10−19≈0.157 mr = \frac{1.67\times10^{-27}\times3\times10^{7}}{1.6\times10^{-19}\times2} = \frac{5.01\times10^{-20}}{3.2\times10^{-19}} \approx 0.157\ \text{m}

The cyclotron (and hence the required alternating-voltage) frequency is …

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