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 will be zero on the y-axis i.e.,
`x=0=z`

magnetic fild cannot be zero in region I and region IV bec-
ause in region I magnetic field is along positive z-direction
due to all the three wires, while in region IV magnetic field
is along negative z-axis due to all the three wires. It can be
zero only in region II and III.

Let magnetic fild be zero on line `(z=0) and x=x`. Then
magnetic fild on this line due to wires (1) and (2) will be
along negative z-axis and due to wire (3) along positive
z-axis. Thus,
`B_1+B_2=B_3`
or `(mu_0)/(2pi) i/(d+x)+(mu_0)/(2pi) i/x=(mu_0)/(2pi) i/(d-x)`
or `1/(d+x)+1/x=1/(d-x)`
This equation gives `x+-d/sqrt3` where magnetic field is
zero.
(b) In this part we change our coordinate axes system just for
butter understanding.

There are three wires (1),(2) and (3) as shown Fig.
If we displace wire (2) toward the z-axis, then force of
attraction per unit length between wires (1 and 2) and (2
and 3) will be given by

`F=(mu_0)/(2pi) (i^2)/r`
The component of F along x-axis will be cancelled out.
Net resultant force will be towards negative z-axis (or mean
position) and will be given by
`F_(n et)=(mu_0)/(2pi) (i^2)/r (2 cos theta)=2{(mu_0)/(2pi) (i^2)/r}z/r=(mu_0)/pi (i^2)/(z^2+d^2) z`
If `zlt ltd, then Z^2+d^2=d^2 and F_(n et)=-((mu_0)/pi (i^2)/(d^2)).z`
Negative sign implies that `F_(n et)` is restoring in nature
Therefore, `F_(n et)prop -z`
i.e., the wire will oscillate simple harmonically.
Let a be the acceleration of wire in this position and `lambda` is
the mass per unit length of this wire then
`F_(n et)=lambda.a=-((mu_0)/pi (i^2)/(d^2)).z or a=-((mu_0i^2)/(pi lambda d^2)).za`
Therefore, frequency of oscillation
`f=1/(2pi)sqrt(a/z)=1/(2pi) i/d sqrt((mu_0)/(pi lambda))=f=i/(2pid)sqrt((mu_0)/(pilambda))`