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A square loop OABCO of side I carries a current i. It is placed as shown figure.Find the magnetic moment of the loop

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Magnetic moment `vecM=ivecA`
`vecA=vec(AB)xxvec(BC)`
`vec(AB)=lcos 30^(@)hati+lsin 30(@)hatk `
`vec(BC)=-lhatj`
`vecA=(l cos 30^(@)hati+l sin 30^(@)hatk)xx(-lhatj)`
`=-I^(2)(sqrt(3))/2hatk+(i^(2))/2hati`
So magnetic moment
`vecM=IvecA=(il^(2))/2[hati-sqrt(3)hatk]`
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