Question

Two species of singly charged positive ions of masses 9.9×10⁻²⁷ kg and 2.7×10⁻²⁶ kg enter a magnetic field at the same location with a speed of 71000 m/s. The charge on the ion is 1.60218× 10⁻¹⁹ C. If the strength of the field is 0.1 T, and they move perpendicularly to the field, find their distance of separation after they complete one half of their circular path. Answer in units of cm.

Answer 1

`d=7.578\,cm`

Step-by-step explanation:

Given:

• mass of the one ion,
  `m_1=9.9* 10^(-27)\,kg`

• mass of another ion,
  `m_2=2.7* 10^(-26)\,kg`

• velocity of projection into the magnetic field,
  `v=71000 \,m.s^(-1)`

• charge on each ion,
  `q_1=q_2=1.60218* 10^(-19)C`

• magnetic field,
  `B=0.1 \,T`

We have, force on a charged particle due to the perpendicular magnetic field:

`F=q.v.B`............................(1)

Force on mass moving in a circular trajectory

`F=(m.v^2)/(r)`...............................(2)

where "r" is the radius of trajectory.

from eq. (1) & (2)

`m.v=q.B.r`

putting the values for the condition 1:

`9.9* 10^(-27)* 71000= 1.60218* 10^(-19)* 0.1* r_1`

`r_1=0.04387\,m`

`r_1=4.387\,cm`

putting the values for the condition 2:

`2.7* 10^(-26)* 71000= 1.60218* 10^(-19)* 0.1* r_2`

`r_2=0.11965\,m`

`r_2=11.965\,cm`

∵The charges have covered a semicircular path from the point of entrance,

∴ their difference in the radius will give their distance from each other.

`d=r_2-r_1`

`d=11.965-4.387`

`d=7.578\,cm`