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A horizontal wire AB which is free to move in a vertical plane and carries a steady current of i_(2) A, is in equilibrium at a height of h over another parallel long wire CD, which is fixed in a horizontal plane and carries a steady current of i_(1) A as shown. Show that when AB is slightly depressed and released it executes simple harmonic motion. Find the period of oscillations. |
Answer» Solution : Since upper wire is in EQUILIBRIUM, for this currents in wires should be UN opposite directions. Magnetic force per UNIT length between wires `(mu_(0)i_(1)i_(2))/(2pi h)`, upwards Let mass of unit length of wire `= lambda` Weight of upper wire per unit length `= lambda g` downward In equilibrium, `(mu_(0)i_(1)i_(2))/(2pi h) = lambda g`...(i) Let `AB`be SLIGHTLY depressed by `x`  Restoring force `F = -[(mu_(0)i_(1)i_(2))/(2pi(h - x)) - lambda g]` `= -[(mu_(0)i_(1)i_(2))/(2pi h(1 - x//h)) - lambda g]` `= -[(mu_(0)i_(1)i_(2))/(2pih)(1 - (x)/(h)) - lambda g]` `= - [lambda g(1 - (x)/(h))^(-1) - lambda g]` `= - lambda g[(1 - (x)/(h))^(-1) - 1]` ` = - lambda g[1 - (x)/(h) - 1] = (lambda g)/(h)x` (since `x ltlt h`) `a = (F)/(lambda) = -(g)/(h)x` `a alpha - x` (CASE of `S.H.M`), `(a = -omega^(2)x)` `T = 2pi sqrt((h)/(g))` |
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