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A circular coil of radius \( 10 \mathrm{~cm} \) and \( 100 \) turns carries a current \( 1 \mathrm{~A} \). What is the magnetic
moment of the coil ?
Options:
moment of the coil ?
Solution:
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Verified Answer
The correct answer is:
\( 3.142 \mathrm{Am}^{2} \)
Given, radius of coil, $r=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$; number of turns, $\mathrm{N}=100$; current, $\mathrm{I}=1 \mathrm{~A}$.
We know, magnetic moment of coil, $\mathrm{M}=\mathrm{NIA}$
Where $\mathrm{A}=$ area $=\pi r^{2}$
Therefore,
$M=N$ inr $^{2}=100 \times 1 \times 3.14 \times\left(10 \times 10^{-2}\right)^{2}=3.142 \mathrm{Am}^{2}$
Magnetic moment of the coil is $3.142 \mathrm{Am}^{2}$
We know, magnetic moment of coil, $\mathrm{M}=\mathrm{NIA}$
Where $\mathrm{A}=$ area $=\pi r^{2}$
Therefore,
$M=N$ inr $^{2}=100 \times 1 \times 3.14 \times\left(10 \times 10^{-2}\right)^{2}=3.142 \mathrm{Am}^{2}$
Magnetic moment of the coil is $3.142 \mathrm{Am}^{2}$
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