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`        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. `
10 years ago

```							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.

```
10 years ago
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