A bar magnet of length 14 cm is placed in the magnetic meridian with… | JEE Main 2021 (16 Mar, Shift 1)

Q6JEE Main 2021 · 16 MAR · SHIFT 1Medium

Earth's Magnetism

A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is: (1G = 10 $-$ 4 T)

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Answer: (b)

N pole towards north: neutral point on the equatorial line, $\frac{\mu_0}{4\pi}\frac{M}{(r^2+l^2)^{3/2}} = B_H$ .
$l = 7$ cm, $r = 18$ cm: $r^2 + l^2 = 373\ \text{cm}^2$ , $(373)^{3/2} \approx 7203\ \text{cm}^3 = 7.2 \times 10^{-3}\ \text{m}^3$
$M = \frac{0.4 \times 10^{-4} \times 7.2 \times 10^{-3}}{10^{-7}} \approx 2.88\ \text{J T}^{-1}$
So the answer is (b) 2.880 J T $^{-1}$ .

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