Two magnets of equal mass are joined at right | vedclass

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Question

(b)For equilibrium of the system torques on $M_1$ and $M_2$ due to $B_H$ must counter balance each other

i.e. ${M_1} 	imes {B_H} = {M_2} 	imes {B

Vedclass · Accepted Answer

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

Vedclass · Answer

(b)For equilibrium of the system torques on $M_1$ and $M_2$ due to $B_H$ must counter balance each other

i.e. ${M_1} 	imes {B_H} = {M_2} 	imes {B_H}$. If $	heta$ is the angle between $M_1$ and $B_H$ will be $(90 - 	heta )$;

so ${M_1}{B_H}\sin 	heta = {M_2}{B_H}\sin (90 - 	heta )$

$ \Rightarrow 	an 	heta = \frac{{{M_2}}}{{{M_1}}} = \frac{M}{{3M}} = \frac{1}{3} \Rightarrow 	heta = {	an ^{ - 1}}\left( {\frac{1}{3}} ight)$

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