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The work done in turning a magnet of mag...

The work done in turning a magnet of magnetic moment 'M' by an angle of `90^(@)` from the meridian is 'n' times the corresponding work done to turn it through an angle of `60^(@)`, where 'n' is given by

Text Solution

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Here, `W_(1)=MB(cos 0^(@)-cos 90^(@))=MB(1-0)=MB`
Similarity, `W_(2)=MB(cos 0^(@)-cos 60^(@))`
`" " MB=(1-(1)/(2))=(MB)/(2)`
`therefore " " W_(1)=2W_(2) Rightarrow n=2`
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