A regular hexagon of side a. A wire of length 24a is coiled on that hexagon. If current in hexagon is I then find the magnetic moment.
A
3√3
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B
6√3Ia2
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C
3√32Ia2
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D
6Ia2
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Solution
The correct option is D6√3Ia2 Let number of turns =n n×6a=24a n=4 Now magnetic moment M=nIA=4.I.A Now area of hexagon 12a2sin(120)+a2asin60+12a2sin(120) =√3a24+√3a2+√3a24=3√3a22 So magnetic moment =4.I.(3√3a22)=6√3Ia2