A proton and a singly charged ion of mass 97 atomic mass units (amu) are acceler

Question

A proton and a singly charged ion of mass 97 atomic mass units (amu) are accelerated through the same potential difference and enter a region of uniform magnetic field moving perpendicular to the magnetic field. What is the ratio of their kinetic energies? I know the ratio is = 1 If the radius of the proton's circular path in the magnetic field is 6.0 cm , what is the radius of the path of the singly charged ion of 97 amu? By what factor must the magnetic field be multiplied by for the singly charge ion to have the same radius path as the proton?

Explanation / Answer

The magnetic field supplies centripetal force, so qvB = mv2/r, and r = mv/qB. q and B are constant, m is 97 times higher for the ion, and v is (mproton/mion) 1/97, since E = mv^2 = constant. Therefore r(ion) will be 97r(proton) = 59.1 cm. To achieve the same radius as the proton, one would need a magnetic field 97 = 9.85 times the strength of the proton's B-field.

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