InterviewSolution
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InterviewSolution
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A thin wire ring of radius `a` carrying a charge `q` approcahes the observation point `P` so that its centre moves rectilinearly with a constant velocity `v`. The plane of the ring remians perpendicular to the motion direction. At what distance `x_(m)` from the point `P` will the ring be located at the moment when the displacement current density at the point `P` becomes maximum? What is the magnitude of this maximum density ? |
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Answer» We have, `E_(P) = (q x)/(4pi epsilon_(0) (a^(2) + x^(2))^(3//2))` then `j_(4) = (del D)/(del t) = epsilon_(0) (del E)/(del t) = (qv)/(4pi (a^(2) + x^(2))^(5//2)) (a^(2) - 2x^(2))` This is maximum, when `x = x_(m) = 0`, and minimum at some other value. The maximum displacement current density is `(j_(d))_(max) = (qv)/(4pi a^(3))` To check this we calculate `(del j_(d))/(del x)`, `(del j_(4))/(del x) = (q v)/(4pi) [(-4x (a^(2) + x^(2)) -5x(a^(2) - 2x^(2))]` This vanishes for `x = 0` and for `x = sqrt((3)/(2)) a`. The latter is easily shown to be a smaller local minimum (negative maximum). |
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