A beam of prop particles (helium nuclei) is used to treat a tumor located 10.0 cm inside a patient. To penetrate to the tumor, the prop particles must be accelerated to a speed of 0.458c, where c is the speed of light. (Ignore relativistic effects?

A beam of prop particles (helium nuclei) is used to treat a tumor located 10.0 cm inside a patient. To penetrate to the tumor, the prop particles must be accelerated to a speed of 0.458c, where c is the speed of light. (Ignore relativistic effects.) The mass of an prop particle is 4.003 u. The cyclotron used to accelerate the beam has radius 1.00 m. What is the magnitude of the magnetic field?

1 Answer
A08
May 26, 2017

2.87T

Explanation:

Lorentz Force equation of charged particleq moving with a velocity v in an Electric field vecE and magnetic field vecB is given as
vecF=q(vecE+vecvxxvecB)

Considering the magnetic field which is to be found out.
In a cyclotron, magnetic field vecB is perpendicular to the velocity vecv of charge q.

Hence, |vecF| = q|vecv||vecB|
=>|vecB|=|vecF| /(q|vecv|) ........(1)

As the alpha particles move in a circular field of radius r these experience centripetal force provided by the magnetic field.
|vecF| = (mv^2)/r .........(2)
substituting value of force from (2) in 91) we get
|vecB| = ((mv^2)/r)/(q|vecv|)
=>|vecB|= (m|vecv|)/(qr)

Using m_alpha = 6.644xx10^-27 kg, |vecv| = 0.458c, r = 1.00 m, and q = 3.204 xx 10^-19 C for "He"^(2+) we get
|vecB|= (6.644xx10^-27xx0.458xx 2.998xx10^8)/(3.204 xx 10^-19xx1.00)
|vecB|= 2.87T

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