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Question

A wire of length \( 2 \, \text{m} \) carrying \( 4 \, \text{A} \) is at \( 60^\circ \) to a magnetic
field of \( 0.25 \, \text{T} \). What is the force on the wire?

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Explanation

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. F = I l B sin θ . F = 4 × 2 × 0.25 × sin 60° = 2 × 0.866 = 1.732 ≈ 1.73 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A

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