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A uniform conducting wire of length is 24 a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side ' a' and then in the form of a square of side ' a '. The coil is connected to a voltage source V_{0}. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1: \sqrt{y} where y is__________.
 

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best_answer

Magnetic moment = NIA
For\, equilateral\, triangle= A_{1}= \frac{1}{2}\times b\times h
                                                                 = \frac{1}{2}\times a\times \frac{\sqrt{3}}{2}a
                                                                 =\frac{\sqrt{3}a^{2}}{4}
h=\frac{\sqrt{3}}{2}a

A_{2}= A_{square}= a^{2}
24a= N_{1}\left ( 3a \right )= N_{2}\left ( 4a \right )         

N_{1}= 8                                            

N_{2}= 6


\frac{M_{eq\cdot triangle}}{M_{square}}= \frac{N_{1}IA_{1}}{N_{2}IA_{2}}
                        = \frac{8}{6}\times \frac{\sqrt{3}a^{2}}{4\times a^{2}}
                         = \frac{1}{\sqrt{3}}
\therefore y= 3

Posted by

vishal kumar

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