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In the arrangement shown in Fig itis possible to measure(by meansof a balance) theforce withwhich a paramagneticball of volumeV = 41 mm^(3) is attratedto a pole of theelectromagnet M.The magnetic induction at the axisof thepoleshoe dependson theheightx as B = B_(0) exp (-ax^(2)), where B_(0) = 1.50 T, a = 100 m^(-2). Find: (a) at what heightx_(m) the ball experiences the maximumattraction, (b) the magnetic suspeptibilityof the paranagneticif the maximumattraction forceequalsF_(max) = 160 muN. |
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Answer» Solution :The force is QUESTION is `vec(F) = (vec(P_(m)) . vec(grad)) vec(B) = (chi BV)/(mu mu_(0)) (dB)/(DX)` since `B` is essentially in the X-direction So, `F_(x) = (chi V)/(2 mu_(0)) (dB^(2))/(dx) = (chi B_(0)^(2) V)/(2 mu_(0)) (d)/(dx) (e^(-2 ax^(2))) = -4 ax e^(-2ax^(2)) (chi B_(0)^(2))/(2 mu_(0)) V` This is maximum when is derivative vanishes i.e., `16 a^(2) x^(2) - 4a = 0`, or, `x_(m) = (1)/(SQRT(4a))` The maximum force is, `F_(max) = 4a (1)/(sqrt(4a)) e^(-1//2) (chi B_(0)^(2) V)/(2 mu_(0)) = (chi B_(0)^(2) V)/(mu_(0)) sqrt((a)/(e))` So, `chi = (mu_(0) F_(max) sqrt((e)/(a)))//V B_(0)^(2) = 3.6xx10^(-4)` |
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