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A particle of mass m and charge q enters a region of electric field vec(E) as shown in the figure with some velocity at point P. At the moment the particle collides elastically with smooth surface at N, the electric field vec(E) is switched off and a magnetic field vec(B) perpendicular to the plane of paper aoutomatically switched on. If the particle hits the surface at point O, then if B=xsqrt((mE)/(qd)). What is the value of x^(1) |
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Answer» `(d)/(2)=(1)/(2)((qE)/(m))xxt^(2)impliest=SQRT(((m)/(qE))d)` Ifu is the velocity while entering the field, then  `d=uxxtimplies u=(d)/(t)=sqrt((qEd)/(m))` also if `upsilon_(N_|_)` is the velocity of the particle in the direction of `vec(E)` at N, then `upsilon_(N_|_)=sqrt(2((qE)/(m))xx(d)/(2))=sqrt((qEd)/(m))` So,net velocity at N is `upsilon_(N)=sqrt(upsilon_(N_|_)^(2)+u^(2))=sqrt((2qEd)/(m))` Since `tan theta-(upsilon_(N_|_))/(u)=1impliestheta=45^(@)` HENCE, the collosion is elastic and the particle is reflected back with the same speed `upsilon_(N)`. The charge q now move with a speed `upsilon_(N)` in MAGNETIC field `vec(B)` . Radius of its path. `R=(m upsilon_(N))/(qB)=(sqrt((2mEd)/(q)))xx(1)/(8) ...................(1)` From figure `R=(d)/(2)xx sqrt(2)=(d)/(sqrt(2))` Putting in equation `(1)` , `(d)/(sqrt(2))=(1)/(B)sqrt((2mEd)/(q))` or `B=2sqrt((mE)/(QD))` |
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