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IBW-898 { ( x ) = ã / m di / dy = - bod / n answer Let me see if I can help you with your problem. I'll tackle it step by way of. **Problem Statement:** Find the magnitude of the magnetic field at the center of a rectangular loop of wire carrying a current of 1000 N. The loop is 0.80 m wide and 0.90 m tall. **Approach:** 1. **Understand the Problem:** - We have a rectangular loop with dimensions 0.80 m (width) and 0.90 m (height). - The loop carries a current of 1000 A. - We need to find the magnetic field at the center of this loop. 2. **Relevant Equations:** - The magnetic field at the center of a rectangular loop due to one side is given by: B = (μ₀ * I) / (2 * R) where μ₀ is the permeability of free space, I is the current, and R is the width of the loop. - However, since we have four sides, we'll calculate the contribution from each side and then add them up. 3. **Calculate the Magnetic Field:** - The loop has two longer sides (0.90 m) and two shorter sides (0.80 m). - The magnetic field at the center due to one of the longer sides is: B₁ = (μ₀ * I) / (2 * 0.90) - The magnetic field at the center due to one of the shorter sides is: B₂ = (μ₀ * I) / (2 * 0.80) - Since there are two long sides and two short sides, the total magnetic field at the center is: B_total = 4 * (B₁ + B₂) / 2 (since each side contributes equally to the total field) - Simplify this: B_total = 2 * (B₁ + B₂) = 2 * [(μ₀ * I) / (2 * 0.90) + (μ₀ * I) / (2 * 0.80)] = 2 * [μ₀ * I (1 / (2 * 0.90) + 1 / (2 * 0.80))] = 2 * [μ₀ * I (1 / 1.80 + 1 / 1.60)] = 2 * [μ₀ * I (0.5556 + 0.625)] = 2 * [μ₀ * I (1.1806)] = 2 * [1.1806 * μ₀ * I] = 2.3612 * μ₀ * I 4. **Plug in the Numbers:** - μ₀ = 4π × 10⁻⁷ Tm/A - I = 1000 A - B_total = 2.3612 * (4π × 10⁻⁷) * 1000 = 2.3612 * 12.5664 × 10⁻⁷ * 1000 = 2.3612 * 12.5664 × 10⁻⁷ * 1000 = 2.3612 * 12.5664 × 10⁻⁴ = 29.6894 × 10⁻⁴ = 0.00296894 T - Convert to mT: B_total = 0.00296894 * 1000 = 2.96894 mT **Final Answer:** The magnitude of the magnetic field at the center of the loop is **2.97 mT**.

30 Nov 2022