Question

A wire of length $1m$ is made into a circular loop and it carries a current of $3.14A$. The magnetic dipole moment of the current loop (in ${Am}^2$) is :

• A. $1$
• B. $0.5$
• C. $0.25$
• D. $0.314$

Answer 1

Answer: (C)

$0.25$

Given: $L=1m$ and $i=3.14A$ $=\pi A$

Radius of the circular loop, $r=\frac{L}{2\pi }$

$\therefore$ $r=\frac{1}{2\pi }$ $m$

Area of the loop, $A=\pi r^2=\pi \times \frac{1}{4{\pi }^2}=\frac{1}{4\pi }$ $m^2$

Dipole moment, $\mu =iA$

$\therefore$ $\mu =\pi \times \frac{1}{4\pi }=0.25$ $A{m}^2$