4. A long horizontal rigidly supported wire carries acurrent of 100 A.

1.	4. A long horizontal rigidly supported wire carries acurrent of 100 A. Directly above it and parallel to it isa fine wire that carries a current of 200 A and weighs0.05 N/ m. How far above the lower wire should thefine wire be kept to support it by magnetic repulsion?Ans. 8 cm.
Answer» Downward force = upward force see attachment, here it is clear that , down ward force is weight = mg and upward force is magnetic force = Bil Hence, magnetic force per unit length = weight per unit distance So, μI₁I₂/2πr /meter length = mg/meter length Here, μ = 4π × 10^-7 , r is the require distance between wires mg/meter length = 0.05N/m , I₁ = 100A , I₂ = 200A ⇒ 4π × 10⁻⁷ × 100A × 200A/2πr = 0.05 ⇒ 2 × 2 × 10⁻³ = 5 × 10⁻²r r = 0.4/5 m = 0.08 m or 8 cm Hence, require distance between wires is 8 cm

4. A long horizontal rigidly supported wire carries acurrent of 100 A. Directly above it and parallel to it isa fine wire that carries a current of 200 A and weighs0.05 N/ m. How far above the lower wire should thefine wire be kept to support it by magnetic repulsion?Ans. 8 cm.

Answer»

Downward force = upward force see attachment, here it is clear that , down ward force is weight = mg and upward force is magnetic force = Bil

Hence, magnetic force per unit length = weight per unit distance So, μI₁I₂/2πr /meter length = mg/meter length Here, μ = 4π × 10^-7 , r is the require distance between wires mg/meter length = 0.05N/m , I₁ = 100A , I₂ = 200A

⇒ 4π × 10⁻⁷ × 100A × 200A/2πr = 0.05 ⇒ 2 × 2 × 10⁻³ = 5 × 10⁻²r r = 0.4/5 m = 0.08 m or 8 cm

Hence, require distance between wires is 8 cm

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