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A bar magnet of length 8 cm is aligned along the magnetic meridian with its north pole directed toward geographic north. On the equatorial line of the magnet, two neutral points are found, and the distance between them is 6 cm. If the horizontal component of Earth’s magnetic field is 3.2 × 10⁻⁵ T, what is the magnetic pole strength of the magnet?

1. 5 ab-amp × cm
2. 10 ab-amp × cm
3. 2.5 ab-amp × cm
4. 20 ab-amp × cm

Correct answer: 5 ab-amp × cm

Solution

Neutral points lie 3 cm from centre (6/2) on the equatorial line, half-length l = 4 cm. Equatorial field M/(r^2+l^2)^1.5 = B_H gives M*1/125 = 0.32 G, so M = 40 G*cm^3. With length 8 cm, pole strength m = M/8 = 5 ab-amp*cm.

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