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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 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30A, as shown in figure . Show that when AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations

Amit Saxena , 10 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
When AB is steady,
Weight per unit length = Force unit length
Weight per unit length = μ0 / 4π 2I1 I2 / r …(i)
NOTE : When the rod is depressed by a distance x, then the force acting on the upper wire increases and behaves as a restoring force
Restoring force / length = μ0 / 4π 2I1 I2 / r
= μ0 / 4π 2I­1 I2 [1 / r – x – 1/ r]
=μ0 / 4π 2I1 I2 x / r(r - x)
When x is small i.e., x << r then r = x ≈ r
Restoring force / length F = μ0 / 4π 2 I­1 I2 / r2 x
Since, F ∝ x and directed to equilibrium position
∴ The motion is simple harmonic
∴ μ0 / 4π 2I1 I2 / r2 = (mass per unit length) ω2 …(ii)
From (i), (Mass per unit length) x g = μ­0 / 4π 2I1 I2 / r
Mass per unit length = μ0 /4π 2I1 I2 / rg … (iii)
From (ii) and (iii)
μ0 / 4 π 2I1 I2 / r2 = μ0 / 4 π 2I1 I2 / rg x ω2 ⇒ ω = √ g / r
⇒ 2π / T = √ g/ r
⇒ T = 2π √ r / g = 2π √ 0.01 / 9.8 = 0. 2 sec
Thanks
Navjot Kalra
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