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

`(i)` We known that magnetic field due to an infinitly long current carryig wire at distance r is given by `B=(mu_(0))/(4pi)((2I)/(r))`. The direction of B uis given by Right hand plam no. 1.
Hence in case of three identical wires resultant field can be zero only if the point P is between the two wire otherwise field B due to all the wires will be in the same direction and so resultant B cannot be zero. Hence, if point P is at a distance x from the central wire s shown in the figure, then
`vec(B)_(p)=vec(B)_(PA)+vec(B)_(PB)+vec(B)_(PC)`
Where `vec(B)_(PA)=` magnetic field at P due to A
`vec(B)_(PB) =` magnetic field at P due to B
`vec(B)_(PC) =` magnetci field at P due to C
`vec(B)_(P)=(mu_(0))/(4pi)2I[(1)/(d+x)+(1)/(x)-(1)/(d-x)](-hat(k))`
For `vec(B)_(P)=0 ,` On solving we get `x=+-d sqrt(3)`
`(ii)` The force per unit length between two parallel current carrying wires is given by
`(mu_(0))/(4pi)(2I_(1)I_(2))/(r)f ("say")`
And is attractive if currents are in the same direction. So when the wire B is displaced along z , the restoring force per unit length `F//l` on the wire B due to wires A and C will be
`(F)/(l)=2fcos theta =2(mu_(0))/(4pi)(2I_(1)I_(2))/(r)xx(z)/(r)["as"cos theta =(z)/(r)]`
or `(F)/(l)=(mu_(0))/(4pi).(4l^(2))/(4pi).(4l^(2))/((d^(2)+z^(2)))z["as"I_(1)=I_(2)=I" and " r^(2)=d^(2)+z^(2)]`
or `(F)/(l)= - (mu_(0))/(4pi).((2I)/(d))^(2)z["as"dgt gtz` and F is opposite to `z] .........(1)`
Since `F prop -z` the motion is simple harmonic .
Comparing equation `(1)` with the standard equation of S.H.M which is
`F=-m omega^(2)z i.e., (F)/(l)-(m)/(l)omega^(2)z=-lambdaomega^(2)z` we get
`lambda^(2)=(mu_(0))/(4pi)xx(4I^(2))/(d^(2))implies omega=sqrt((mu_(0)I^(2))/(pi d^(2)lambda))implies2pin`
`=(I)/(d)sqrt((m_(0))/(pilambda))impliesn=(1)/(2pid)sqrt((mu_(0))/(pilambda))`