A short magnet of magnetic moment 6.75 , Am^2 | vedclass

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(C) At the neutral point, the magnetic field produced by the magnet is equal in magnitude and opposite in direction to the horizontal component of the

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$30$

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(C) At the neutral point, the magnetic field produced by the magnet is equal in magnitude and opposite in direction to the horizontal component of the Earth's magnetic field.For a point on the axial line of a short bar magnet, the magnetic field $B$ is given by:$B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$Given:Magnetic moment $M = 6.75 \, Am^2$Horizontal component of Earth's magnetic field $B_H = 5 	imes 10^{-5} \, Wb/m^2$$\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$Equating the fields:$10^{-7} 	imes \frac{2 	imes 6.75}{d^3} = 5 	imes 10^{-5}$$\frac{13.5 	imes 10^{-7}}{d^3} = 5 	imes 10^{-5}$$d^3 = \frac{13.5 	imes 10^{-7}}{5 	imes 10^{-5}} = 2.7 	imes 10^{-2} = 0.027 \, m^3$$d = \sqrt[3]{0.027} = 0.3 \, m$Converting to centimeters: $d = 0.3 	imes 100 = 30 \, cm$.

$A$ short magnet of magnetic moment $6.75 \, Am^2$ produces a neutral point on its axis. If the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, Wb/m^2$, then the distance of the neutral point from the center of the magnet is.....$cm$.

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