Practice question
Question
Two parallel wires \( 0.11 \, \text{m} \) apart carry currents of \( 6 \, \text{A} \) and \( 2 \,
\text{A} \) in the same direction. What is the force per unit length between them? (\( \mu_0 = 4 \pi
\times 10^{-7} \, \text{T m/A} \))
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. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 6 × 2/2 π × 0.11) = (48 × 10⁻⁷/0.22) = 2.18 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.