Question

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

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

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