A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20A, is in equilibrium at a height of 0.01m over another parallel wire CD which is fixed in a horizontal plane. Show that AB is slightly depressed , it executes simle harmonic motion. Find the period of oscillations.

Weight per unit length of wire is balanced by magnetic force per unit length.

or, { 4¶*10^-7*i₁*i₂ }/d² = mg/l       where i₁  =   Current through wire CD

                                                                      i₂  =    Current through wire AB     =   20A

                                                                      d=      Vertical separation between wires   = 0.01 mtrs.

Let wire AB is depressed by 'x' then,

                                                               Vertical separation between wires = d-x

Now magnetic force per unit length on AB =   { 4¶*10^-7*i₁*i₂ }/(d-x)² = K/(d-x)²   where K is a constant equal to  4¶*10^-7*i₁*i₂.

or,   Net vertical force per unit length on AB =[  { 4¶*10^-7*i₁*i₂ }/(d-x)² ] - (mg/l) = K [1/ (d-x)² - 1/d² ]

                                                                             =  K [( 2dx-x² )/{ (d-x)² *d² } ]  

Taking x/d=µ  &  x<<d;i.e. µ<<1

Simplifying above expression give R.H.S.  =  2dx/d⁴ [(1-µ/2)/{(1-µ)²}]K = 2x/d³ K        : Ignoring the negligible terms 1-µ & 1-µ/2 equal to 1.

L.H.S  = (m/l)w²x = 2x/d*( K/d² )   ; or,    w² = 2g/d          where w=angular frequency of SHM with F=constant*x(Hookes' Law)

                                                                                                   K/d² = mg/l

So,period = 2¶/w= ( 2¶ ) / [ (2g/d)^1/2 ] = 0.14 sec.