A bar magnet, having a length of 1 cm and a cross-sectional area of 1.0 cm^2 , generates a magnetic field of 1.5 × 10^ -4 T at an end-on position located 15 cm from its centre. Determine the magnetic moment M of the

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Correct answer: C. 2.5 A m^2

The magnetic field at an axial (end-on) position of a short bar magnet is given by:

B = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3}

Since the length of the magnet

l = 1 \text{ cm}

is much smaller than the distance

d = 15 \text{ cm}

, the short magnet approximation is valid.

Substituting the given values:

1.5 \times 10^{-4} = 10^{-7} \times \dfrac{2M}{(0.15)^3}

M = \dfrac{1.5 \times 10^{-4} \times (0.15)^3}{2 \times 10^{-7}}

M = \dfrac{1.5 \times 10^3 \times 0.003375}{2}

M = \dfrac{5.0625}{2}

M \approx 2.5 \text{ A m}^2

A bar magnet, having a length of 1 cm and a cross-sectional area of...

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