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A uniform conducting wire of length length 10a and resistance R is wound up into four turn as a current carrying coil in the shape of equilateral triangle of side a. If current I is flowing 4 through the coil then the magnetic moment of the coil is A. `(sqrt3)/(2)a^(2)I`B. `(a^(2)I)/sqrt3`C. `sqrt3a^(2)I`D. `(2a^(2)I)/(sqrt3)` |
Answer» Correct Answer - C Magnetic moment M = NIA `=4xxIxxa^(2)(sqrt3)/(4)=sqrt3a^(2)I` |
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