Question

A U-shaped electromagnet shown in Figure is designed to lift a 400 kg mass (which includes the mass of the keeper). The iron yoke ($\mu_r$ = 3000) has a cross section of 40 $cm^2$ or and mean length of 50 cm, and the air gaps are each 0.1 mm long. Neglecting the reluctance of the keeper, calculate the number of turns in the coil when the excitation current is 1 A.

Answer 1

324

Answer 2

To calculate the number of turns in the coil, we need to determine the required magnetic field strength (B) to lift the given mass and then use the magnetic circuit analogy (Ampere's Law) to find the necessary magnetomotive force (MMF), from which the number of turns can be derived.

1. Convert all given parameters to SI units:

• Mass to be lifted, $M = 400 \text{ kg}$
• Relative permeability of iron yoke, $\mu_r = 3000$
• Cross-sectional area of yoke, $A = 40 \text{ cm}^2 = 40 \times (10^{-2})^2 \text{ m}^2 = 40 \times 10^{-4} \text{ m}^2 = 4 \times 10^{-3} \text{ m}^2$
• Mean length of iron yoke, $l_m = 50 \text{ cm} = 0.5 \text{ m}$
• Length of each air gap, $l_g = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 1 \times 10^{-4} \text{ m}$
• Excitation current, $I = 1 \text{ A}$
• Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ H/m}$
• Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

2. Calculate the force required to lift the mass: The force required to lift the mass is equal to its weight. $F = M \times g$ $F = 400 \text{ kg} \times 9.8 \text{ m/s}^2 = 3920 \text{ N}$

3. Relate the magnetic force to the magnetic field (B) in the air gap: The magnetic force exerted by an electromagnet on a ferromagnetic material (keeper) is given by $F = \frac{B^2 A}{2\mu_0}$ for one pole. Since the U-shaped electromagnet has two poles acting on the keeper, the total force is the sum of forces from both poles. $F_{total} = 2 \times \frac{B^2 A}{2\mu_0} = \frac{B^2 A}{\mu_0}$ We need to find the magnetic field B: $B^2 = \frac{F_{total} \mu_0}{A}$ $B = \sqrt{\frac{F_{total} \mu_0}{A}}$ Substitute the values: $B = \sqrt{\frac{3920 \text{ N} \times 4\pi \times 10^{-7} \text{ H/m}}{4 \times 10^{-3} \text{ m}^2}}$ $B = \sqrt{3920 \times \pi \times 10^{-4}}$ $B = \sqrt{3.92 \times \pi \times 10^{-1}}$ $B = \sqrt{1.2315}$ (approximately, using $\pi \approx 3.14159$) $B \approx 1.1097 \text{ T}$

4. Apply Ampere's Circuital Law (Magnetic Circuit Analogy) to find the number of turns (N): The total magnetomotive force (MMF), $NI$, is the sum of the MMF drops across the iron yoke and the two air gaps. $NI = H_m l_m + 2 H_g l_g$ Where $H_m$ is the magnetic field intensity in the iron yoke and $H_g$ is the magnetic field intensity in the air gap. We know that $B = \mu H$, so $H = B/\mu$. For the iron yoke, $\mu_m = \mu_r \mu_0$, so $H_m = \frac{B}{\mu_r \mu_0}$. For the air gap, $\mu_g = \mu_0$, so $H_g = \frac{B}{\mu_0}$. Substitute these into the MMF equation: $NI = \frac{B}{\mu_r \mu_0} l_m + 2 \frac{B}{\mu_0} l_g$ Factor out $B/\mu_0$: $NI = \frac{B}{\mu_0} \left( \frac{l_m}{\mu_r} + 2l_g \right)$ Now, substitute the known values: $N \times 1 \text{ A} = \frac{1.1097 \text{ T}}{4\pi \times 10^{-7} \text{ H/m}} \left( \frac{0.5 \text{ m}}{3000} + 2 \times 1 \times 10^{-4} \text{ m} \right)$ $N = \frac{1.1097}{4\pi \times 10^{-7}} \left( 0.00016667 + 0.0002 \right)$ $N = \frac{1.1097}{12.566 \times 10^{-7}} \left( 0.00036667 \right)$ $N = (8.830 \times 10^5) \times (3.6667 \times 10^{-4})$ $N \approx 323.7$

Rounding to the nearest whole number, the number of turns $N \approx 324$.