Let's solve Problem P.6-52 step by step. **Given:** - Total mean length of flux path in iron, \( l_{Fe} = 3~\text{m} \) - Yoke-bar contact area, \( A = 0.01~\text{m}^2 \) - Relative permeability of iron, \( \mu_r = 4000 \) - Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7}~\text{H/m} \) - Each air gap, \( l_g = 2~\text{mm} = 0.002~\text{m} \) - **Assume there are two air gaps** (since the problem says "each"). - Mass to be lifted, \( m = 100~\text{kg} \) - Acceleration due to gravity, \( g = 9.81~\text{m/s}^2 \) --- ### **Step 1: Find the Force Required** The force required to lift the mass is: \[ F = m \cdot g = 100 \times 9.81 = 981~\text{N} \] --- ### **Step 2: Relate Force to Magnetic Field** The force produced by the electromagnet across the air gap is: \[ F = \frac{B^2}{2\mu_0} \cdot A \] Solving for \( B \): \[ B = \sqrt{\frac{2 F \mu_0}{A}} \] Substitute values: - \( F = 981~\text{N} \) - \( \mu_0 = 4\pi \times 10^{-7}~\text{H/m} \) - \( A = 0.01~\text{m}^2 \) \[ B = \sqrt{ \frac{2 \times 981 \times 4\pi \times 10^{-7}}{0.01} } \] Calculate numerator: \[ 2 \times 981 = 1962 \] \[ 1962 \times 4\pi \times 10^{-7} = 1962 \times 12.566 \times 10^{-7} \approx 24645.6 \times 10^{-7} = 2.46456 \times 10^{-3} \] Now divide by 0.01: \[ \frac{2.46456 \times 10^{-3}}{0.01} = 0.246456 \] Now take the square root: \[ B = \sqrt{0.246456} \approx 0.496~\text{T} \] --- ### **Step 3: Find the Magnetomotive Force (MMF) Needed** The total MMF required is: \[ \text{MMF} = H_{Fe} l_{Fe} + 2 H_g l_g \] But since the permeability of iron is much higher than air, most of the MMF drops across the air gaps. We'll check this below, but let's be precise. #### **Find Flux, \(\Phi\):** \[ B = \frac{\Phi}{A} \implies \Phi = B \cdot A = 0.496 \times 0.01 = 0.00496~\text{Wb} \] #### **Field Strength in Air Gap:** \[ B = \mu_0 H_g \implies H_g = \frac{B}{\mu_0} \] \[ H_g = \frac{0.496}{4\pi \times 10^{-7}} \approx \frac{0.496}{1.25664 \times 10^{-6}} \approx 3.95 \times 10^5~\text{A/m} \] Total length of air gap: \( 2 \times 0.002 = 0.004~\text{m} \) Total MMF across air gaps: \[ \text{MMF}_{gap} = H_g \cdot (2l_g) = 3.95 \times 10^5 \times 0.004 = 1580~\text{A-turns} \] #### **Field Strength in Iron:** \[ B = \mu_0 \mu_r H_{Fe} \implies H_{Fe} = \frac{B}{\mu_0 \mu_r} \] \[ H_{Fe} = \frac{0.496}{4\pi \times 10^{-7} \cdot 4000} = \frac{0.496}{5.0265 \times 10^{-3}} \approx 98.7~\text{A/m} \] Total length in iron: \( l_{Fe} = 3~\text{m} \) Total MMF in iron: \[ \text{MMF}_{iron} = H_{Fe} \cdot l_{Fe} = 98.7 \times 3 = 296~\text{A-turns} \] --- ### **Step 4: Total MMF Needed** \[ \text{MMF}_{total} = \text{MMF}_{iron} + \text{MMF}_{gap} = 296 + 1580 = 1876~\text{A-turns} \] --- ## **Final Answers:** ### **Step-by-step summary:** 1. **Required force:** \( F = 981~\text{N} \) 2. **Required flux density:** \( B = 0.496~\text{T} \) 3. **Flux:** \( \Phi = 0.00496~\text{Wb} \) 4. **MMF in air gaps:** \( 1580~\text{A-turns} \) 5. **MMF in iron:** \( 296~\text{A-turns} \) 6. **Total MMF needed:** \(\boxed{1876~\text{A-turns}}\) --- **Final Answer:** \[ \boxed{1876~\text{A-turns}} \] is the total MMF needed to lift a 100 kg mass with the given electromagnet.