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(C) The neutral point for a magnet placed with its south pole pointing towards the geographic north lies on its axial line. The magnetic field due to

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$1200 \, ab-amp \cdot cm^2$

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(C) The neutral point for a magnet placed with its south pole pointing towards the geographic north lies on its axial line. The magnetic field due to the magnet at an axial point is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2Mr}{(r^2 - l^2)^2}$. For a short magnet,this simplifies to $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$.At the neutral point,the magnetic field of the magnet equals the horizontal component of the Earth's magnetic field $(B_H)$:$\frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} = B_H$Given: $r = 20 \, cm = 0.2 \, m$,$B_H = 0.3 	imes 10^{-4} \, T$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.Substituting the values:$10^{-7} \cdot \frac{2M}{(0.2)^3} = 0.3 	imes 10^{-4}$$10^{-7} \cdot \frac{2M}{0.008} = 0.3 	imes 10^{-4}$$2M = 0.3 	imes 10^{-4} 	imes 0.008 	imes 10^7$$2M = 0.3 	imes 8 = 2.4$$M = 1.2 \, A \cdot m^2$.Since $1 \, A \cdot m^2 = 1000 \, ab-amp \cdot cm^2$,we have:$M = 1.2 	imes 1000 = 1200 \, ab-amp \cdot cm^2$.

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