Three infinitely long thin wires, each carrying current `i` in the same direction, are in the `x-y` plane of a gravity free space . The central wire is along the `y - axis` while the other two are along `x = +-d`.
(i) Find the locus of the points for which the magnetic field `B` is zero.
(ii) If the central wire is displaced along the ` Z- direction` by a small amount and released, show that it will excecute simple harmonic motion . If the linear density of the wires is `lambda`, find the frequency of oscillation.

Explanation

(a)
Magnetic field is zero on middle wire i.e. y-axis i.e. `x = 0`.
Let magnetic field is zero at point `P`
`(mu_(0)i)/(2pi) ((1)/(d + x) + (1)/(x)) = (mu_(0)i)/(2pi(d - x))`
`((2x + d))/(x(d + x)) = (1)/((d - x))`
`(2x + d)(d - x) = x(d + x)`
`2xd - 2x^(2) + d^(2) -xd = xd + x^(2)`
`x^(2) = (d^(2))/(3) rArr x = +- (d)/(sqrt(3))`
Locus of points where magnetic field is zero
`x = 0, x = +- (d)/(sqrt(3))`
(b)
When middle wire is displaced along z-axis, force on it due to other wire
`F = (mu_(0)i.i)/(2pisqrt(a^(2) + z^(2)))`
Restoring force on middle wire
`F' = - 2F cos theta - 2. (mu_(0)i^(2))/(2pi sqrt((d^(2)+ z^(2)))).(z)/(sqrt((d^(2) + z^(2)))`
`= (mu_(0)i^(2)z)/(pi(a^(2) + z^(2)))`, downward
Acceleration of middle wire
`a = (F')/(lambda) = (mu_(0)i^(2)z)/(pi(d^(2) + z^(2))lambda)`
Since `z lt ltd, z^(2)` is negligible
`a = -((mu_(0)i^(2))/(pi lambda d^(2)))z`
`a alpha - z` (case of `S.H.M.`)
`a = -omega^(2)z`
`omega = sqrt((mu_(0)i^(2))/(pi lambda d^(2)))`
`f = (omega)/(2pi) = sqrt((mu_(0)i^(2))/(4pi^(3) lambdad^(2)))`
`= (i)/(2pi d)sqrt((mu_(0))/(pi lambda))`
Note: `F': fo rce//l eng th, lambda: mass//l eng th`.