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Question:
Published on: 3 December, 2024

Derive an expression for lifting power of a magnet

Answer:

Consider two poles arranged directly opposite each other as shown in Fig. 29. Let each have an area A square meters and F Newtons be the force of alteration between them. Let one pole be moved by an incremental distance \(\delta x\) against the force Fj then the work done is obviously F\(\delta x\) Juoules.

The volume of the magnetic field between the poles is increased by A\(\delta x\)m3

Fig.29

Hence the energy stored increased by

Energy density of the magnetic field \(\times\)A\(\delta x\) Joules

Or \(F\delta x=\frac{1}{2}\left(\frac{B^2A}{\mu_0\mu}\delta x\right)Joules\)

Energy is stored in the air gap.

Hence \(\mu=1\)

\(F=\frac{B^2A}{{2\mu}_0}\)

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