Lifting Power of a Magnet: Concept, Formula & Solved Examples - SSC JE Electrical | Magnetic Circuit

In the blog of electromagnetism, understanding how much weight a magnet can lift is a fundamental topic. The lifting force depends on the strength of the magnetic field and the area of the magnet's pole faces. This article explains the lifting power of a magnet using a step-by-step derivation and real-world examples.

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What is Lifting Power of a Magnet?

The lifting power or pulling force of a magnet refers to the maximum force it can exert to hold or lift a ferromagnetic object. This force is directly related to the magnetic flux density (B) and the pole area (A) of the magnet.

Derivation of Lifting Force of a Magnet

Let’s consider a magnet with pole area A (in m²) and pulling force P (in newtons) between the poles. If one of the poles is pulled apart by a small distance dx (in metres), the work done is:

$$\text{Work done}=P\cdot dx\text{(i)}$$

This work goes into creating an additional magnetic field volume:

$$\text{Additional volume} = A \cdot dx$$

The energy density of a magnetic field is:

$$\frac{B^2}{2{\mu }_0}\text{ (Joules per cubic metre)}$$

So, the total energy required is:

$$\text{Energy}=\frac{B^2}{2{\mu }_0}\cdot A\cdot dx\text{(ii)}$$

Equating equations (i) and (ii):

$$P\cdot dx=\frac{B^2}{2{\mu }_0}\cdot A\cdot dx$$$$\Rightarrow P=\frac{B^2\cdot A}{2{\mu }_{\mathrm{0}}}$$

Thus, the formula for lifting force becomes:

$$\boxed{{\displaystyle P=\frac{B^2\cdot A}{2{\mu }_0}\text{(in Newtons)}}}$$

Where:

• P = Pulling force (N)
• B = Magnetic flux density (Wb/m²)
• A = Pole face area (m²)
• μ₀ = Permeability of free space

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Alternate Units of Lifting Power

In terms of pressure or force per unit area:

$$\frac{P}{A} = \frac{B^2}{2\mu_0} \quad \text{(N/m²)}$$

Or in terms of kilogram-force (kgf):

$$P=\frac{B^2\cdot A}{2{\mu }_0\cdot 9.81}\text{(in kgf)}$$

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Example 1: Horse-shoe Magnet Lifting a Load

Problem:
A horse-shoe magnet is made from wrought iron of length 45.7 cm and cross-sectional area 6.45 cm². Two exciting coils of 500 turns are placed on each limb. Calculate the exciting current needed to lift a 68 kg load.
Given:

• Relative permeability ​=700
• Negligible reluctance of the load
• Close contact between magnet and load

Solution:

Load per pole = 68 / 2 = 34 kg
Force per pole = 34 × 9.81 = 333.5 N
A = 6.45 cm² = 6.45 × 10⁻⁴ m²

Using the formula:

$$P=\frac{B^2\cdot A}{2{\mu }_0}\Rightarrow 333.5=\frac{B^2\cdot 6.45\times {10}^{-4}}{2\cdot 4\pi \times {10}^{-\mathrm{7}}}$$

Solving for B:

$$B\approx 1.14\text{Wb/m²}$$

Now, calculate magnetic field strength (H):

$$H=\frac{B}{{\mu }_0{\mu }_r}=\frac{1.14}{4\pi \times {10}^{-7}\times 700}\approx 1296\text{AT/m}$$

Length of iron = 0.457 m
Total ampere-turns (AT) required = $H \cdot l = 1296 \times 0.457 = 592.6$

Total turns = 2 × 500 = 1000 turns
Exciting current = $\frac{592.6}{1000}=\boxed{{\displaystyle 0.593\text{A}}}$

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Example 2: Electromagnet Lifting a 1000 kg Ingot

Problem:
An electromagnet with a pole face area of 0.5 m² per pole must lift an iron ingot of 1000 kg. The air gap between the pole and ingot is 1 mm.
Assume:

• Relative permeability of iron = infinity

Solution:

Iron offers zero reluctance, so force is applied only across air gap.

Using:

$$P = \frac{B^2 \cdot A}{\mu_0} \Rightarrow 1000 \cdot 9.8 = \frac{B^2 \cdot 0.5}{4\pi \times 10^{-7}} \Rightarrow B = 0.157 \, \text{Wb/m²}$$

Now,

$$H = \frac{B}{\mu_0} = \frac{0.157}{4\pi \times 10^{-7}} \approx 125 \times 10^3 \, \text{AT/m}$$

Total length of air gap = 2 × 1 mm = 2 × 10⁻³ m

AT required = $H \cdot l = 125 \times 10^3 \cdot 2 \cdot 10^{-3} = \boxed{250 \, \text{AT}}$

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Conclusion

Understanding the lifting power of magnets is crucial in designing electromagnets for industrial applications. The key takeaway is the relation:

$$P = \frac{B^2 \cdot A}{2\mu_0}$$

This helps engineers determine the current, magnetic field strength, and design parameters to achieve desired lifting capacity.

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FAQs on Lifting Power of a Magnet

Q1. What factors affect the lifting power of a magnet?

• Magnetic flux density (B), pole area (A), and the permeability of the medium between poles.

Q2. Why is the permeability of iron assumed infinite in some problems?

• To simplify calculations, assuming infinite permeability means iron offers zero reluctance to magnetic flux.

Q3. What is μ₀ in magnetism?

• It’s the permeability of free space,

Q4. How can I increase the lifting power of an electromagnet?

• A: By increasing the current, number of turns, or using a material with higher permeability.