# Three infinitely long thin wires, each carrying current i  in the same direction , are in the x –y 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 of 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, show that it will execute simple harmonic motion. If the linear density of the wires is λ, find the frequency of oscillation.

Navjyot Kalra
10 years ago
Hello Student,
(i) KEY CONCEPT : Magnetic field due to an infinitely long current carrying wire at distance r is given by
B = μ0 / 4π (2i / r)
The direction B is given by right hand palm rule.
Hence, in case of three identical wires, resultant field can be zero only if the point P is between the two wires, otherwise field B due to all the wires will be in the same direction and so resultant B cannot be zero. Hence, if point P is at a distance x from the central wire as shown in figure, the,
(ii) KEY CONCEPT : The force per unit length between two parallel current carrying wires is given by μ0 / 4π 2i1 i2 /r = f (say)
And is attractive if currents are in the same direction.
So, when the wire B is displaced along Z – axis by a small distance Z, the restoring per unit length F / ℓ on the wire B due to wires A and C will be
F / ℓ = 2 f cos θ = 2 μ0 / 4π 2i1 i2 / r x z / r [ ass cos θ = z / r]
or F / ℓ = μ0 / 4π 4i2 z / (d2 + z2) [as I1 = I2 and r2 = d2 + z2]
or F / ℓ = μ0 / 4π (2i / d)2 z [as d >> z and F is opposite to z] …..(I)
Since F ∝ - z, the motion is simple harmonic.
Comparing eq. (1) with the standard equation of S.H.M which is
F = - mω2 z i.e., F / ℓ = - m / ℓ ω2 z
= - λ ω2 z , we get
λ ω20 4π x 4i2 / d2 ⇒ ω = √ μ0 i2 / πd2λ
⇒ 2π n = i / d √ μ0 / πλ ⇒ n = i = / 2πd √μ0 / πλ
Thanks
Navjot Kalra