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Question: The dipole moment of a short bar magnet is 1.25 A-m<sup>2</sup>. The magnetic field on its axis at a...

The dipole moment of a short bar magnet is 1.25 A-m2. The magnetic field on its axis at a distance of 0.5metre from the centre of the magnet is

A
B

4×10−24 \times 10 ^ { - 2 } Newton / amp - metre

C
D

6.64×10−86.64 \times 10 ^ { - 8 } Newton / amp - metre

Answer
Explanation

Solution

B=μ04π2Md3=10−7×2×1.25(0.5)3=2×10−6 N/A−mB = \frac { \mu _ { 0 } } { 4 \pi } \frac { 2 M } { d ^ { 3 } } = 10 ^ { - 7 } \times \frac { 2 \times 1.25 } { ( 0.5 ) ^ { 3 } } = 2 \times 10 ^ { - 6 } \mathrm {~N} / \mathrm { A } - \mathrm { m }