Three infinitely long thin wires, each carrying cu

Question

Three infinitely long thin wires, each carrying current iin 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 · Accepted Answer

Hello Student,Please find the answer to your question(i) KEY CONCEPT : Magnetic field due to an infinitely long current carrying wire at distance r is given byB = &mu;0 / 4&pi; (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 &mu;0 / 4&pi; 2i1 i2 /r = f (say)And is attractive if currents are in the same direction.So, when the wire B is displaced along Z &ndash; axis by a small distance Z, the restoring per unit length F / ℓ on the wire B due to wires A and C will beF / ℓ = 2 f cos θ = 2 &mu;0 / 4&pi; 2i1 i2 / r x z / r [ ass cos θ = z / r]or F / ℓ = &mu;0 / 4&pi; 4i2 z / (d2 + z2) [as I1 = I2 and r2 = d2 + z2]or F / ℓ = &mu;0 / 4&pi; (2i / d)2 z [as d >> z and F is opposite to z] &hellip;..(I)Since F &prop; - z, the motion is simple harmonic.Comparing eq. (1) with the standard equation of S.H.M which isF = - m&omega;2 z i.e., F / ℓ = - m / ℓ &omega;2 z= - &lambda; &omega;2 z , we get&lambda; &omega;2 =&mu;0 4&pi; x 4i2 / d2 &rArr; &omega; = √ &mu;0 i2 / &pi;d2&lambda;&rArr; 2&pi; n = i / d √ &mu;0 / &pi;&lambda; &rArr; n = i = / 2&pi;d √&mu;0 / &pi;&lambda;ThanksNavjot KalraaskIITians Faculty

Magnetism> Three infinitely long thin wires, each ca...

Shane Macguire , 10 Years ago

Navjyot Kalra

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Magnetism> Three infinitely long thin wires, each ca...

Shane Macguire , 10 Years ago

Navjyot Kalra

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