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A researcher studying the properties of ions in the upper atmosphere wishes to construct an apparatus with the following charateristics: Using an electric field, a beam of ions, each having charge q, mass m , and initial velocity vhati, is turned through an angle of 90^@ as each ion undergoes displacement Rhati + Rhatj. The ions enter a chamber as shown in figure and leave through the exit port with the same speed they had when they entered the chamber. The electric field acting on the ions is to have constant magnitude. If the field is produced by two flat plates and is uniform in direction, what value should the field have in this case? |
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Answer» `(mv^2)/(2QR) (hati+hatj)` `qE=(mv^(2))/(R)` or `E=(mv^(2))/(qR)` As final velocity along x-axis becomes zero and finally the ions STARTS moving towards y direction the electric field should have component towards x-and y-directions, respectively. for x-component of electric field (using `v_(x)^(2)=u_(x)^(2)+2a_(x)trianglex)` `0=V^(2)-2((qE_(x))/(m))R` or `E_(x)=(mv^(2))/(2qR)` for y-component of electric field (again using `v_(y)^(2)=u_(y)^(2)+2atriangley)` `v^(2)=0+2((qE_(y))/(m))R` or `E_(y)=(mv^(2))/(2qR)` ltbr. Hence net electric field is `vecE=-(mv^(2))/(2qR)hati+(mv^(2))/(2qR)hatj=(mv^(2))/(2qR)(-hati+hatj)` |
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