Question

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?

Answer 1

`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`