A long horizontal wire `AB`,which is free to move in a vertical plane and carries a steady current of `20 A`, is in equilibrium at a height of `0.01 m` over another paralllel long wire `CD` which is fixed in a horizontal plane and carries a steady current of `30 A`, as shown in figure. Show that when `AB` is slightly
depressed it executes simple harmonic motion. Find the period of oscillations.

Answer

0.2

Explanation

Let m be mass per unit length of wire AB. At a height x above the wire CD, magnetic force per unit length on wire AB will be given by:
`F_(m)=(mu_(0))/(2pi)(i_(1)i_(2))/(x^(2)).dx` (upwards)
Weight per unit length of wire AB is
`F_(g)=mg` (downwards)
Here, m = mass per unit length of wire AB
At x = d, wire is in equilibrium i.e.
`F_(m)=F_(g)`
Or `(mu_(0))/(2pi)(i_(1)I_(2))/d=mg` or `(mu_(0))/(2pi)(i_(1)i_(2))/(d^(2))=(mg)/d`
When AB is depressed, x decreases therefore, `F_(m)` will increase, while `F_(g)` remains the same. Let AB is displaced by dx downwards. Differentiating Eq (i) w.r.t. x, we get
`d_(m)=-(mu_(0))/(2pi)(i_(1)i_(2))/(x^(2)). dx`..............iii
i.e. restoring force `F=dF_(m)prop -dx`
Hence the motion of wire is simple harmonic
From Eqs. (ii) and (iii), we can write
`dF_(m)=-((mg)/d).dx:.` Acceleration of wire `a=-(g/d).dx`
Hence period of oscillation
`T=2pi sqrt(|(dx)/a|)=2pisqrt((|"displacement"|)/(|"acceleration"|))` or `T=2pisqrt(d/g)=2pisqrt(0.01/9.8)` or `T=0.2s`