Question

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the ion enters a magnetic field of 0.522 T in a direction perpendicular to the field. The charge on the ion is 1.602 × 10−19 C. Find the radius of the ion’s path in the field. Answer in units of cm.

Answer 1

1.8 cm

Step-by-step explanation:

`m` = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

`q` = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

`\Delta V` =Potential difference through which the ion is accelerated = 215 V

`v` = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

`(0.5) m v^(2) = q \Delta V\\(0.5) (3.46*10^(-26)) v^(2) = (1.6*10^(-19)) (215)\\(1.73*10^(-26)) v^(2) = 344*10^(-19)\\v = 4.5*10^(4) ms^(-1)`

`r` = Radius of the path followed by ion

`B` = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

`qvB = (mv^(2) )/(r) \\qB = (mv)/(r)\\r =(mv)/(qB)\\r =((3.46*10^(-26))(4.5*10^(4)))/((1.6*10^(-19))(0.522))\\r = 0.018 m \\r = 1.8 cm`