Question

Three infinitely long thein wires, each carrying curren i int eh same diecrtion, are the dxy plane of a gravilty 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 smal 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 1

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

Magnetic field cannot be zero in region I and egion IV because in reion 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 be zero only i reginn II and III.
Let magnetic field is zero on line `z=0` and `x=x` (shown as dotted). The magnetic field on the line due to wire 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_0i)/(2pix)=mu_0/(2pi) i/((d-x))`
`or 1/(d+x)+1/x=1/(d-x)`
this equation gives `x=+-d/sqrt3`
Hence there will be two lilnes `x=d/sqrt3`
and `x=-d/sqrt3 (z=0)`
where magnetic field is zero.
ii. In this part we chasnge our coordinate axes sytem, just for better understanding.

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

`f=mu_0/(2pi) i^2/r`
The componetns of `F` along x-axis will be canceled out. Net resultant force will be towards negative z-axis (or mean position) and will be given by
`F_("net")=2Fcostheta=2{mu_0/(2pi)i^2/r}z/r`
`F_("net")=mu_0/pi i^2(z^2+d^2).z(r^2=z^2+d^2)`
If `zltltd` then
`z^2d^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` the mass per unit length of wire, then
`F_("net") =lamda a-(mu_0/pii^2/d^2)z`
`or ap=((m_0i^2)/(pilamdad^2))z`
`:. `Frequency of oscillation
`f=1/(2pi) sqrt(|"acceleration")/("displacement"|)`
`1/(2pi) sqrt(|a/z|=1/(2pi) i/d sqrt(m_0/(pilamda))`
or `f=1/(2pid) sqrt(mu_0)/(pilamda))`