A short bar magnet with its north pole facing | vedclass

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(D) Initially,the neutral point is obtained on the equatorial line of the magnet. At the neutral point,the magnitude of the horizontal component of th

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$\sqrt{5} B_H$

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(D) Initially,the neutral point is obtained on the equatorial line of the magnet. At the neutral point,the magnitude of the horizontal component of the Earth's magnetic field $(B_H)$ is equal to the magnetic field due to the bar magnet on its equatorial line $(B_e)$. Thus,$|B_H| = |B_e|$.When the magnet is rotated by $90^o$,the point $P$ now lies on the axial line of the magnet. The magnetic field due to the bar magnet on its axial line is $B_a = 2 B_e$. Since $B_e = B_H$,we have $B_a = 2 B_H$.The Earth's horizontal magnetic field $B_H$ remains in the same direction. The net magnetic field $B$ at point $P$ is the vector sum of $B_a$ and $B_H$,which are perpendicular to each other.Therefore,the net magnetic induction is $B = \sqrt{B_a^2 + B_H^2} = \sqrt{(2 B_H)^2 + B_H^2} = \sqrt{4 B_H^2 + B_H^2} = \sqrt{5} B_H$.

$A$ short bar magnet with its north pole facing north forms a neutral point at $P$ in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane,the net magnetic induction at $P$ is (Horizontal component of earth's magnetic field = $B_H$)

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