Magnet Pull Force Calculator, Formula, Magnet Pull Force Calculation

Magnet Pull Force Calculator:

Enter the values of permeability of the medium between the poles, μ­(H/M), q1(Am) and q2(Am) are the magnetic pole strengths of the magnets and distance between the centers of the two poles, r(m) to determine the value of magnet pull force, F(N).

Enter Permeability of the Medium: H/M
Enter Magnetic Pole Strengths of the Magnets: Am
Enter Magnetic Pole Strengths of the Magnets: Am
Enter Distance Between the Center of Two Poles: s
   
Result – Magnet Pull Force: N

Magnet Pull Force Formula:

Magnetic pull force is a measure of the attractive or repulsive force between two magnetic poles. The strength of this force depends on the magnetic properties of the materials, the distance between them, and the medium through which they interact.

Magnet pull force, F(N) in Newton is calculated by dividing the product of the permeability of the medium between the poles, μ­(H/M) in Henries per metre, q1(Am) and q2(Am) are the magnetic pole strengths of the magnets in Ampere-metre by the value of 4pi and the distance between the centers of the two poles, r(m) in metres.

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Magnet pull force, F(N) = μ­(H/M) * q1(Am) * q2(Am) / 4 * 3.14 * r2(m)

F(N) = magnet pull force in Newton, N.

μ­(H/M) = permeability of the medium in Henries per metre, H/M. (4 * 10-7 H/M)

q1(Am) & q2(Am) = magnetic pole strengths of the magnets in Ampere-metres, Am.

r(m) = distance between the center of two poles in metres, m.

Magnet Pull Force Calculation:

  1. Suppose two small magnets have pole strengths of 1×10-2 Am and 2×10-2 Am, respectively, and are placed 0.05 metres apart in air.

Given: q1(Am) = 1×10-2, q2(Am) = 2×10-2Am, μ­(H/M) = 4 * 10-7 H/M, r(m) = 0.05 m

Magnet pull force, F(N) = μ­(H/M) * q1(Am) * q2(Am) / 4 * 3.14 * r2(m)

F(N) = 4 * 10-7 * 1×10-2 * 2×10-2 / 4 * 3.14 * 0.052

F(N) = 8 * 10-9N.

  1. The force between two point charges is given, along with the permittivity of the medium, the value of one charge, and the distance between them. We want to find the value of the second charge, q2.
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Given: q1(Am) = 5×10-2, F(N) = 2 * 10-7 N, μ­(H/M) = 4 * 10-7 H/M, r(m) = 0.1 m

Magnet pull force, F(N) = μ­(H/M) * q1(Am) * q2(Am) / 4 * 3.14 * r2(m)

q2(Am) = F(N) * 4 * 3.14 * r2(m) / μ­(H/M) * q1(Am)

q2(Am) = 02 * 10-7 * 4 * 3.14 * 0.12 / 4 * 10-7 * 5 * 10-2

q2(Am) = 2 * 10-1Am.

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