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A wire of length 314 \mathrm{~cm} carrying current of 14 \mathrm{~A} is bent to form a circle. The magnetic moment of the coil is_________ \mathrm{A-} \mathrm{m}^{2}. [Given \pi=3.14]

Option: 1

11


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{L=314 \mathrm{~cm}=2 \pi R ;\: \: R=50 \mathrm{~cm}}

\mathrm{I=14 \mathrm{~A}}

\mathrm{M= Magnetic\: moment\: of\: the\: coil =I A}

                                                                   \mathrm{=14 \times \pi(0.5)^2 }

                                                                    \mathrm{=\frac{14}{4} \times \frac{22}{7} }

\mathrm{M =11 A-m^2}



 

Posted by

Sanket Gandhi

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