Three infinitely long thin wires, each carrying current i in the same dieretion, are the xy 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 execute simple harmonic motion. If the linear mass density of the wires is `lamda`, find the frequency of oscillation.

Answer

(i) 3 (ii) 1

Explanation

(i)3 (ii) 1
(i) Magnetic field will be zero on the y-axis i.e.
`x=o=z`

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

Let magnetic field is zero on line (z = 0) and x = x. Then magnetic field 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)1/(d+x)+(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 `zx=+-d/(sqrt(3))`, where magnetic field is zero.
(ii) In this part, we change our coordinate axes system, just for better understanding.

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

The components 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_("net")=(mu_(0))/(2pi)(i^(2))/(2pi)(i^(2))/r(2 cos theta)={(mu_(0))/(2pi) (i^(2))/r}z/r`
`F_("net")=(mu_(0))/(pi)(i^(2))/((z^(2)+d^(2)))z`
if ` z lt lt d`,then `, z^(2)+d^(2)=d^(2)` and `F_("net")=-((mu_(0))/(pi) (i^(2))/(d^(2))).z`
Negative sign implies that `F_("net")` is restoring in nature
Therefore `F_("net")prop -z`
i.e. the wire will oscillate simple harmonically.
Let a be the acceleration of wire in this position and `lamda` is the mass per unit length of this wire then
`F_("net")=lamda,a=-((mu_(0))/(pi) (i^(2))/(d^(2)))z` or `a=-((mu_(0)i^(2))/(pi lamda d^(2))).z:.` Frequency of oscillation
`f=1/(2pi)sqrt(("acceleration")/("displacement"))=1/(2pi)sqrt(a/z)=1/(2pi)i/d sqrt((mu_(0))/(pi lamda))` or `f=i/(2pid) sqrt((mu_(0))/(pi lamda))`