Practice question
Question
A wire of length \( 1.3 \, \text{m} \) carrying \( 6 \, \text{A} \) is at \( 45^\circ \) to a magnetic
field of \( 0.8 \, \text{T} \). What is the force on the wire?
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 = I l B sin θ . F = 6 × 1.3 × 0.8 × sin 45° = 6.24 × 0.707 = 4.4117 ≈ 4.41 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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