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Find the magnitude of magetic moment of the current carrying lop `ABCDEFA`. Each side of the loop is `10 cm` long and current in the loop is `i=2.0A`

Text Solution

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By assuming two equal and opposite currents in `BE`, two current carrying loops (`ABEFA` and `BCDEB`) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence,

`M_("net")=sqrt(M^2+M^2)=sqrt2M`
where `M=iA=(2.0)(0.1)(0.1)=0.02A-m^2`
`M_("net")=(sqrt2)(0.02)A-m^2`
`=0.028A-m^2`
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