A cyclotron has oscillatory frequency of `3 xx 10^(7)` Hz and a dee radius of 60 cm. A deuteron is accelerated in the given cyclotron. Find the magnitude of magnetic field used to accelerate the deuteron. Also, calculate the energy possessed by the deutron after coming out from the cyclotron (in MeV).
Take, mass of deutron ` = 3.34 xx 10^(-27)` kg

Explanation

Deutron is an isotope of hydrogen consisting of a proton and neutron in its nucleus.
`:.` Charge on deutron - charge on a proton ` = 1.6 xx 10^(-19)C`
Given, oscillation frequency of cyclotron, `f = 3 xx 10^(7)` Hz
Radius of deer `r = 60 cm = 0.6` m
Mass of deutron ` = 3.34 xx 10^(-27)` kg
If B is the magnitude of magnetic field inside the cyclotron, then cyclotron frequency is given by
` f = (qB)/(2 pi m) `
` rArr " " B = (f (2 pi m))/q = (3xx 10^(7) xx 2 xx 3.14 xx 10^(-27))/(1.6 xx 10^(-19))`
` = 3.93 ` T
Kinetic energy of the deutron will be
` K = (q ^(2)B^(2)r^(2))/(2 m) `
` = ((1.6 xx 10^(-19))^(2) xx (3.93)^(2) xx (0.6)^(2))/(2 xx 3.34 xx 10^(-27)) `
` = 2.13 xx 10^(-11) J = (2.13 xx 10^(-11))/(1 .6 xx 10^(-19)) eV `
` = 1.33 xx 10^(8) eV`