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)

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
$M=NiA,$ where $A=\pi r^{2}$
$M=Ni.\pi r^{2}$
$=100\times 1\times 3.142\times \left(10 \times \left(10\right)^{- 2}\right)^{2}$
$=100\times 3.142\times 100\times 10^{- 4}$
$=3.142 \, A \, m^{2}$