KCET · Physics · Alternating Current
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 \( 10^{4} \mathrm{Am}^{2} \)
- C \( 3.142 \mathrm{Am}^{2} \)
- D \( 3 A \mathrm{~m}^{2} \)
Answer & Solution
Correct Answer
(C) \( 3.142 \mathrm{Am}^{2} \)
Step-by-step Solution
Detailed explanation
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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