Circular motion of ions in a magnetic field

Electricity and Magnetism Level 2

A beam of ions, each of charge $-3.20 \times 10^{-19}\text{ C}$ and travelling at a speed of $1.2 \times 10^5\text{ m s}^{-1}$ , enters a region where a uniform magnetic field of flux density $2.4 \times 10^{-2}\text{ T}$ acts normally to the original direction of the beam, as shown below.

When the ions are in the magnetic field, they are deflected and they travel in a circular path of radius $0.390\text{ m}$ . What is the nucleon number of the nuclei of the ions in the beam?

4 8 30 15

1 solution

Vincent Moroney
Jun 15, 2018

The beam of ions undergoes circular motion upon entering the perpendicular magnetic field so we know $qv\beta = m\frac{v^2}{r}$ so all we need to do is solve for $m$ and evaluate. Doing this gives $m = \frac{qr\beta}{v} = 2.496\cdot10^{-26}\text{kg}.$ Now to find the nucleon number we divide this number by the atomic mass unit $\mu = 1.67\cdot10^{-27}\text{kg}$ this gives us $\frac{m}{\mu} = \frac{2.496\cdot10^{-26}\text{kg}}{ 1.67\cdot10^{-27} \text{kg}} = \boxed{15}$

Circular motion of ions in a magnetic field

1 solution

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