A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is `9.0xx10^(-5)Vm^(-1)`, make a simple guess as to what the beam contains. Why is the answer not unique ?

Explanation

1. Here, velocity of charged particle is `vecv _|_vecE _|_vecB`. In this situation, for an extremely thin beam of charged particles to pass undeflected through a given region of electric and magnetic field, speed of each particle should be,
`v=E/B=(9xx10^(5))/0.75=1.2xx10^(6)m/s`
2. Now, kinetic energy gained by above particle while passing through a potential difference V is,
`1/2mV^(2)=Vq`
`thereforeq/m=v^(2)/(2V)`
`thereforeq/m=(1.2xx10^(6))/((2)(15xx10^(3)))`
`thereforeq/m=4.8xx10^(7)C/(kg)`
3. Here charged particles in the given beam may be deuterons because for deuterons `(""_(1)H^(2))`,
`("charge")/("mass")=e/(2m_(p))`
(`because` Deuteron in the nucleus which contains one proton and one neutron)
= `(1.6xx10^(-19))/(2xx1.67xx10^(-27))`
= `4.79xx10^(7)C/(kg)`
= `4.8xx10^(7)C/(kg)" "...(1)`
(4) Above answer is not the only answer (or unique answer) because here the charged particles may be,
(i) `He^(+2)` ions for which `("charge")/("mass")=(2e)/(4m_(p))=e/(2m_(p))`
(ii) `Li^(+3)` ions for which `("charge")/("mass")=(3e)/(6m_(p))=e/(2m_(p))`
(iii) `Be^(+4)` ions for which `"charge"/"mass"` = `(4e)/(8m_(p))=e/(2m_(p))`
Answer obtained in equation (1) is not the unique answer.