Question

Four current carrying wires are Placed at the corners of a rectangle as shown in the bring Find the magnitude at net magnetic torce per unit length on the bottom right wire of 12 A 5A 7cm 15A 10 A 15cm​

Answer 1

Disclaimer:

The current carrying wire image is attached.

Concept:

Force due to a  current carrying current can be defined as

F = (µ₀I₁I₂)/(2πr)

where, I₁, I₂ are current on wires and r is the distance between them.

Given:

The four current carrying wires.

Find:

The magnitude of magnetic force per unit length at the bottom wire.

Solution:

Force per unit length due to the 5A wire,

F₁ = (µ₀15 × 5)/(2π×0.07)

F₁ = 2.143 × 10⁻⁴ j N/m

Force per unit length due to the 10 A wire,

F₂ = (µ₀15 × 10)/(2π×0.15)

F₂ = 2 × 10⁻⁴ i N/m

Force per unit length due to the 12 A wire,

F₃ = (µ₀15 × 12)/(2π×√(0.07²+0.15²)) [0.15 i - 0.07 j/√(0.07²+0.15²)]

F₃ = (1.971 × 10⁻⁴ i - 0.920 × 10⁻⁴ j)N/m

Total force, F = F₁ + F₂ + F₃ = (3.971 × 10⁻⁴ i + 1.223 × 10⁻⁴ j )N/m

F = 4.155 ×10⁻⁴ N/m

Hence, the total magnetic force per unit length on the bottom right wire is 4.155 ×10⁻⁴ N/m.

#SPJ2