Q172 / 180 · KCET 2014 · Physics

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 ?

  • A
    \( 3.142 \times 10^{4} \mathrm{~A} \mathrm{~m}^{2} \)
  • B
    \( 10^{4} \mathrm{Am}^{2} \)
  • Correct answer
    C
    \( 3.142 \mathrm{Am}^{2} \)
  • D
    \( 3 A \mathrm{~m}^{2} \)

Solution

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 $3.142 \mathrm{Am}^{2}$

Hence, the correct answer is option: 'C'.

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