Question

A fire door electromagnet is wound with a copper coil of 500 turns. The air gap has a length of 2mm and a cross sectional area of 1000mm2 . A current of 3A is used to energise the electromagnet.

Calculate the force of the magnetic pull.

Note: The reluctance of the iron core can be considered negligible when compared to the air gap. Magnetic fringing and leakage may also be ignored.

Answer 1

The force of the magnetic pull is approximately 70.8 Newtons.

Step-by-step explanation:

The force of the magnetic pull can be calculated using the formula F = (N * I)^2 * μ₀ * A / (2 * g^2), where F is the force, N is the number of turns in the coil, I is the current, μ₀ is the permeability of free space (4π × 10^−7 T·m/A), A is the cross-sectional area of the air gap, and g is the length of the air gap.

Plugging in the values given, we get F = (500 * 3)^2 * (4π × 10^−7) * 0.001 / (2 * 0.002^2), which simplifies to F = 70.8 N.

Therefore, the force of the magnetic pull is approximately 70.8 Newtons.