A bar magnet is placed with its north pole po | vedclass

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(D) Initially,the magnet is placed such that its north pole points north. At a point $P$ on the equatorial line,the magnetic field due to the magnet $

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$\sqrt{5}\,B_H$

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(D) Initially,the magnet is placed such that its north pole points north. At a point $P$ on the equatorial line,the magnetic field due to the magnet $(B_e)$ is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$. Thus,$|B_e| = |B_H|$.When the magnet is rotated by $90^\circ$,the point $P$ now lies on the axial line of the magnet. The magnetic field due to the magnet at an axial point is $B_a = 2B_e = 2B_H$.The direction of the axial magnetic field $B_a$ is perpendicular to the horizontal component of the Earth's magnetic field $B_H$.The resultant magnetic field $B$ at point $P$ is given by $B = \sqrt{B_a^2 + B_H^2}$.Substituting $B_a = 2B_H$,we get $B = \sqrt{(2B_H)^2 + B_H^2} = \sqrt{4B_H^2 + B_H^2} = \sqrt{5B_H^2} = \sqrt{5}\,B_H$.

List-$I$	List-$II$
$(A)$ Magnetic induction	$(I)$ $\text{Ampere meter}^2$
$(B)$ Magnetic intensity	$(II)$ $\text{Weber}$
$(C)$ Magnetic flux	$(III)$ $\text{Gauss}$
$(D)$ Magnetic moment	$(IV)$ $\text{Ampere/meter}$

$A$ bar magnet is placed with its north pole pointing towards the geographic north. The magnetic field at a point $P$ on its equatorial line is zero. If the magnet is rotated by $90^\circ$,what will be the magnetic field at point $P$? The horizontal component of the Earth's magnetic field is $B_H$.

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