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
May 26, 2017

Answer:

#2.87T#

Explanation:

Lorentz Force equation of charged particle#q# 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#