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`(mv^2)/(2qR) (hati+hatj)``(mv^2)/(2qR) (-hati+hatj)``(2mv^2)/(qR) (hati+hatj)``(2mv^2)/(qR) (-hati+hatj)`

Answer :

BSolution :

Concentric cylindrical electrodes will produce radial electric field. As ion is entering at O and leaving at B, hence the path followed by ion should be circular and centrered at A. Required centripetal force should be provided by force on the ion due to electric field. Hence <br> `qE=(mv^(2))/(R)` or `E=(mv^(2))/(qR)` <br> 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)` <br> `0=v^(2)-2((qE_(x))/(m))R` or `E_(x)=(mv^(2))/(2qR)` <br> for y-component of electric field (again using `v_(y)^(2)=u_(y)^(2)+2atriangley)` <br> `v^(2)=0+2((qE_(y))/(m))R` or `E_(y)=(mv^(2))/(2qR)` ltbr. Hence net electric field is <br> `vecE=-(mv^(2))/(2qR)hati+(mv^(2))/(2qR)hatj=(mv^(2))/(2qR)(-hati+hatj)`