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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 ? |
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Answer» Solution :1. Here, velocity of charged particle is `vecv _|_vecE _|_vecB`. In this situation, for an extremely thin beam ofcharged 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. |
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