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Question

A wire of length \( 0.8 \, \text{m} \) carrying \( 3 \, \text{A} \) makes an angle of \( 30^\circ \)
with a magnetic field of \( 0.6 \, \text{T} \). What is the force on the wire?

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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. Force F = I l B sin θ . F = 3 × 0.8 × 0.6 × sin 30° = 2.4 × 0.6 × 0.5 = 0.72 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ

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