The axial field (B_{A}) and the equatorial fi | vedclass

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Question

(A) For a short bar magnet of magnetic moment $m$ at a distance $r$ from its center:$1$. The magnetic field on the axial line is given by: $B_{A} = \f

Vedclass · Accepted Answer

$B_{A} = 2 B_{E}$

Vedclass · Answer

(A) For a short bar magnet of magnetic moment $m$ at a distance $r$ from its center:$1$. The magnetic field on the axial line is given by: $B_{A} = \frac{\mu_{0}}{4 \pi} \frac{2m}{r^{3}}$$2$. The magnetic field on the equatorial line is given by: $B_{E} = \frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}$$3$. Comparing the two expressions,we can see that $B_{A} = 2 	imes \left( \frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}} ight) = 2 B_{E}$.Therefore,the correct relation is $B_{A} = 2 B_{E}$.

The axial field $(B_{A})$ and the equatorial field $(B_{E})$ due to a short bar magnet at equal distances are related as:

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