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Grade upto college level Magnetism

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

Question image for A long horizontal wire AB, which is free to move i

Amit Saxena

12 Years agoGrade upto college level

1 Answer

Navjyot Kalra

12 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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