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Asked in GUJCET 2009 · Field of a bar magnet and earth's field
Idea: for a short bar magnet the axial (end-on) field at distance z from the centre is ⃗B=(μ₀)/(4π)(2⃗M)/(z³).
The moment ⃗M points from the S pole to the N pole inside the magnet, and the whole factor (μ₀)/(4π)2/(z³) is positive.
So on the axis the field is parallel to ⃗M: the lines leave the N pole and continue outwards along the axis.
On the equatorial line the same magnet gives -(μ₀)/(4π)(⃗M)/(z³), which is antiparallel to ⃗M - the contrast worth remembering.
So the axial field is along the direction of the magnetic dipole moment.
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