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Question

Two parallel wires \( 0.07 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) 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

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 8 × 6/2 π × 0.07) = (192 × 10⁻⁷/0.14) = 1.3714 × 10⁻⁵ ≈ 1.37 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =

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