Question

A long horizontal rigidly supported wire carries a current of 100A. Directly above it and parallel to it is a fine wire that carries a current of 200A and weighs 0.05N/m.How far above the lower wire should the fine wire be kept to support it by magnetic repulsion??

Answer 1

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

Answer 2

Answer:

8cm should be the distance between the wire to hold the upper thin wire