Two long parallel wires are separated by 18 cm. One of the wires carries a current of 8 A and the other carries a current of 26 A. The permeabilty of free space is 1.257 × 10−6 N · m/A. Determine the magnitude of the magnetic force on a 2.1 m length of the wire carrying the greater current.

The magnitude of the magnetic force on the wire carrying greater current is is 4.85 x 10⁻⁴ N

Explanation:

Given;

distance of separation of the two wires, d = 18 cm

current in the first wire, I₁ = 8 A

current in the second wire, I₂ = 26 A

length of the wire, L = 2.1 m

permittivity of free space, μ₀ = 4π x 10⁻⁷ N·m/A = 1.257 × 10⁻⁶ N·m/A

The magnitude of the magnetic force on the wire carrying greater current is given as;

Therefore, the magnitude of the magnetic force on the wire carrying greater current is is 4.85 x 10⁻⁴ N

ok

explanation:

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