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A bar magnet of length 3 cm has points A and B along its axis at distances of 24 cm and 48 cm on the opposite sides. Ratio of magnetic fields at these points will be

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$\frac{1}{2 \sqrt{2}}$

answer is A.

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Detailed Solution

Magnetic field due to a bar magnet at a distance r from the centre of magnet on axial position is

$B = \frac{μ_{0}}{4 π} \cdot \frac{2 M}{r^{3}}$

$∴ \frac{B_{1}}{B_{2}} = {(\frac{r_{2}}{r_{1}})}^{3} = {(\frac{48}{24})}^{3} = \frac{8}{1} = 8$

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