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An electron is travelling with a constant velocity vecv=v_(0)hatj in the presence of a uniform electric field vecE=E_(0)hatk, where E_(0)=(mg)/e (Here, m denotes the mass of the electron and e denotes its charge g denotes the acceleration due to gravity). Gravity acts in the -Z direction. At t=0, when the electron is at the origin, a uniform magnetic fiedl vecB=B_(x)hati+B_(y)hatj is switched on. Here B_(x) and B_(y) are fixed positive constants and B_(0)=sqrt(B_(x)^(2)+B_(y)^(2)). The coordinates of the point whre the electron intersects the first time after t=0 are given by: |
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Answer» `((2pimv_(0))/e((B_(x)^(2))/(B_(0)^(3))),(2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))))` Let the angle made by the magnetic field vector with the Y-axis be `theta`. So `(B_(x))/(B_(y))=tan theta` The magnetic field is given by `vecB=(B_(0)sin theta)hati+(B_(0)cos theta)hatj` The electron will follow a helical PATH. The component of velocity of the electron parallel to the field will REMAIN unchanged, and the components of its velocity in a plane perpendicular to the field will oscillate. At `t=0` component of velocity parallel to field `v_(y)=v_(0)cos theta` And, component of velocity perpendicular to field, `v_(_|__)=v_(0)sin theta` So radius of the helical path `r=(mv_(_|_))/(eB_(0))` And time perido `T=(2pim)/(eB_(0))` The electron will return to the X-Y plane when it has completed one full rotation, i.e. at `t=T=(2pim)/(eB_(0))` During this time, the displacement of the electron (in a DIRECTION parallel to the field), `L=v_(y)((2pim)/(eB_(0)))=(2pimv_(0)cos theta)/(eB_(0))` , coordinates of the point where the electron intersects the X-Y plane are ` (L sin theta, L cos theta)` i.e. `((2pi mv_(0)sin theta cos theta)/(eB_(0)), (2pi m v_(0)cos^(2) theta)/(eB_(0)))`, i.e. `(((2pi m v_(0))/e)((B_(x)B_(y))/(B_(0)^(3))),((2pimv_(0))/e)((B_(y)^(2))/(B_(0)^(3))))` |
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