A short bar magnet placed in a horizontal pla | vedclass

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Question

(A) The magnetic field on the axis of the magnet at a distance $d_{1} = 14 \; cm$ is given by:$B_{1} = \frac{\mu_{0}}{4 \pi} \frac{2M}{d_{1}^{3}} = H

Vedclass · Accepted Answer

$11.1$

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(A) The magnetic field on the axis of the magnet at a distance $d_{1} = 14 \; cm$ is given by:$B_{1} = \frac{\mu_{0}}{4 \pi} \frac{2M}{d_{1}^{3}} = H \dots (1)$Where $M$ is the magnetic moment and $H$ is the horizontal component of the Earth's magnetic field.When the magnet is turned by $180^o$,the neutral points shift to the equatorial line of the magnet.The magnetic field on the equatorial line at a distance $d_{2}$ is given by:$B_{2} = \frac{\mu_{0}}{4 \pi} \frac{M}{d_{2}^{3}} = H \dots (2)$Equating $(1)$ and $(2)$:$\frac{2}{d_{1}^{3}} = \frac{1}{d_{2}^{3}}$$d_{2}^{3} = \frac{d_{1}^{3}}{2}$$d_{2} = \frac{d_{1}}{\sqrt[3]{2}} = \frac{14}{1.26} \approx 11.1 \; cm$The new null points will be located at $11.1 \; cm$ on the equatorial line.

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