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(B) For the equilibrium of the system,the torques on magnet $1$ $(M_1)$ and magnet $2$ $(M_2)$ due to the horizontal component of the Earth's magnetic

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$	an^{-1}\left(\frac{1}{3}ight)$

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(B) For the equilibrium of the system,the torques on magnet $1$ $(M_1)$ and magnet $2$ $(M_2)$ due to the horizontal component of the Earth's magnetic field $(B_H)$ must balance each other.Let $	heta$ be the angle that magnet $1$ makes with the magnetic meridian. Since the magnets are joined at a $90^{\circ}$ angle,magnet $2$ will make an angle of $(90^{\circ} - 	heta)$ with the magnetic meridian.The torque on a magnetic dipole in a magnetic field is given by $	au = MB_H \sin \alpha$,where $\alpha$ is the angle between the magnetic moment and the field.For equilibrium,the magnitudes of the torques must be equal:$M_1 B_H \sin 	heta = M_2 B_H \sin(90^{\circ} - 	heta)$Given that $M_1 = 3M_2$,we substitute this into the equation:$3M_2 B_H \sin 	heta = M_2 B_H \cos 	heta$Dividing both sides by $M_2 B_H \cos 	heta$:$3 	an 	heta = 1$$	an 	heta = \frac{1}{3}$$	heta = 	an^{-1}\left(\frac{1}{3}ight)$

List-$I$	List-$II$
$(A)$ Magnetic field intensity	$(i)$ $Wb$
$(B)$ Magnetic flux	(ii) $Wb \cdot m^{-2}$
$(C)$ Magnetic pole strength	(iii) $A \cdot m$
$(D)$ Magnetic induction	(iv) $A \cdot m^{-1}$

Two magnets of equal mass are joined at right angles to each other as shown. Magnet $1$ has a magnetic moment $3$ times that of magnet $2$. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium,what angle will magnet $1$ subtend with the magnetic meridian?

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