Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
The greater current is 0.4 A
Since the two long wires are parallel, the magnetic force, F on each wire is given by
F = μ₀I₁I₂L/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₁ = current in first wire, I₂ = current in second wire, L = length of section of wires = 3.0 m and d = separation distance of the wires = 6.0 mm = 6.0 × 10⁻³ m.
Given that F = 8.0 μN = 8.0 × 10⁻⁶ N and I₂ = 2I₁ (the current in one wire is twice the current in the other wire), we have
F = μ₀I₁I₂L/2πd
F = μ₀ 2I₁I₁L/2πd
F = μ₀I₁²L/πd
Since we require the current in the wire, we make I₁ subject of the formula.
So, I₁ = √(Fπd/μ₀L)
Substituting the values of the variables into the equation, we have
I₁ = √(Fπd/μ₀L)
I₁ = √[8.0 × 10⁻⁶ N × π × 6.0 × 10⁻³ m/(4π × 10⁻⁷ H/m × 3.0 m)]
I₁ = √[48.0π × 10⁻⁹ Nm/12π × 10⁻⁷ H]
I₁ = √[4 × 10⁻² Nm/H]
I₁ = 2 × 10⁻¹ A
I₁ = 0.2 A
Since I₂ is the greater current and I₂ = 2I₁,
I₂ = 2 × 0.2 A
I₂ = 0.4 A
So, the greater current is 0.4 A
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