# 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 $\propto$ particles (helium nuclei) is used to treat a tumor located 10.0 cm inside a patient. To penetrate to the tumor, the $\propto$ particles must be accelerated to a speed of 0.458c, where c is the speed of light. (Ignore relativistic effects.) The mass of an $\propto$ 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?

May 26, 2017

$2.87 T$

#### Explanation:

Lorentz Force equation of charged particle$q$ moving with a velocity $v$ in an Electric field $\vec{E}$ and magnetic field $\vec{B}$ is given as
$\vec{F} = q \left(\vec{E} + \vec{v} \times \vec{B}\right)$

Considering the magnetic field which is to be found out.
In a cyclotron, magnetic field $\vec{B}$ is perpendicular to the velocity $\vec{v}$ of charge $q$.

Hence, $| \vec{F} | = q | \vec{v} | | \vec{B} |$
$\implies | \vec{B} | = | \vec{F} \frac{|}{q | \vec{v} |}$ ........(1)

As the $\alpha$ particles move in a circular field of radius $r$ these experience centripetal force provided by the magnetic field.
$| \vec{F} | = \frac{m {v}^{2}}{r}$ .........(2)
substituting value of force from (2) in 91) we get
$| \vec{B} | = \frac{\frac{m {v}^{2}}{r}}{q | \vec{v} |}$
$\implies | \vec{B} | = \frac{m | \vec{v} |}{q r}$

Using ${m}_{\alpha} = 6.644 \times {10}^{-} 27 k g$, $| \vec{v} | = 0.458 c$, $r = 1.00 m$, and $q = 3.204 \times {10}^{-} 19 C$ for ${\text{He}}^{2 +}$ we get
$| \vec{B} | = \frac{6.644 \times {10}^{-} 27 \times 0.458 \times 2.998 \times {10}^{8}}{3.204 \times {10}^{-} 19 \times 1.00}$
$| \vec{B} | = 2.87 T$