Two identical short magnets are separated by the same centre-to-centre distance r.
In arrangement I, their centres and magnetic moments are collinear. In arrangement II, the line joining their centres is perpendicular to both magnetic moments, which are parallel to each other.
The ratio of the magnitudes of interaction forces $F_I:F_II$ is:
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Question

Two identical short magnets are separated by the same centre-to-centre distance r.

In arrangement I, their centres and magnetic moments are collinear. In arrangement II, the line joining their centres is perpendicular to both magnetic moments, which are parallel to each other.

The ratio of the magnitudes of interaction forces $F_I:F_II$ is:

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G1 Physics · Accepted Answer

For collinear dipoles, | U I ∣ U_I| U I ​ ∣ = (μ₀/4π)·2M²/r³. Therefore, F I F_I F I ​ = |-d U I / d r ∣ U_I/dr| U I ​ / d r ∣ = (μ₀/4π)·6M²/r⁴. So, F I F_I F I ​ = 3μ₀M²/(2πr⁴). For two parallel dipoles on an equatorial line, | U I I ∣ U_II| U I ​ I ∣ = (μ₀/4π)·M²/r³. Hence, F I I F_II F I ​ I = (μ₀/4π)·3M²/r⁴ = 3μ₀M²/(4πr⁴). Therefore, F I / F I I F_I/F_II F I ​ / F I ​ I = 2. Thus, F I : F I I F_I:F_II F I ​ : F I ​ I = 2:1.