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A magnetic field vec(B)=-B_(0)hat(i) exists within a sphere of radius R=v_(0)Tsqrt(3) where T is the time period of one revolution of a charged particle starting its motion form origin and moving with a velocity vce(v)_(0) = (v_0)/(2) sqrt(3) hat(i) - v_(0)/(2) hat(j). Find the number of turns that the particle will take to come out of the magnetic field. |
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Answer» `v_(|/|) =(v_(0)sqrt(3))/(2)` will contributed to helical path. `T=2 pi (m)/(B_(0)Q)` Pitch, `P=v_(|/|)T=(v_(0)sqrt(3))/(2)T` Number of turns `= (R)/(P)=(v_()Tsqrt(3)(2))/(v_(0)sqrt(3)T) =2`. |
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