Question

A circular coil of radius $10 \,cm$ and $100$ turns carries a current $1\,A$. What is the magnetic moment of the coil ?

• A. $3.142 \times 10^4 \, A \, m^2$
• B. $10^4 \, A \, m^2$
• C. $3.142 \, A \, m^2$
• D. $3 \,A\,m^2$

Answer 1

Answer: (C)

$3.142 \, A \, m^2$

Answer 2

Given,
Radius of coil $r=10 \times 10^{-2} m$
Number of turns $N=100$ turn, current in coil $i=1 A$
Magnetic moment $M=?$
By magnetic dipole moment of current loop
$M =N i A,$ where $A=\pi r^{2}$
$M =N i \cdot \pi r^{2}=100 \times 1 \times 3.142 \times\left(10 \times 10^{-2}\right)^{2}$
$=100 \times 3.142 \times 100 \times 10^{-4}=3.142\, A - m^{2}$