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Question

Two parallel wires \( 0.12 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 3 \, \text{A} \) in
the same direction. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A}
\))

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Explanation

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 7 × 3/2 π × 0.12) = (84 × 10⁻⁷/0.24) = 3.5 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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