From the radius to the q/m ratio

Problem

A particle in a cyclotron traces a circle of radius $r = 0{,}5$ m in a field $B = 1{,}2$ T at $v = 4\cdot 10^7$ m/s. Find $q/m$ and compare it with that of the proton.

Solution

From circular motion in a magnetic field $r = \dfrac{mv}{qB}$ the charge-to-mass ratio is obtained: $\frac{q}{m} = \frac{v}{rB} = \frac{4\cdot10^7}{0{,}5\cdot 1{,}2}$ $\ev{q/m \approx 6{,}7\cdot 10^7\;\text{C/kg}}$

For the proton one has $\dfrac{q}{m} = \dfrac{1{,}6\cdot10^{-19}}{1{,}67\cdot10^{-27}} \approx 9{,}6\cdot10^7$ C/kg. The value found is of the same order and not too far off: the particle is similar to a proton, or to a light ion.

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Topics: Magnetismo Concepts: Moto circolare uniforme · Forza di Lorentz