Question

A wire of length L metre carrying current I ampere is bent in the form of a circle. What is the magnitude of magnetic dipole moment?

• A. $I{L}^2/4\pi$
• B. $I^2{L}^2/4\pi$
• C. $I^{2}L/8\pi$
• D. $I{L}^2/8\pi$

Answer 1

The correct option is A $I{L}^2/4\pi$

Current carrying loop behaves as magnetic dipole.
Consider a plane loop of wire carrying current. Looking at the upper face, current is anticlockwise. Therefore, it has a north polarity. Looking at the lower face of the loop, current is clockwise. Therefore, it has a south polarity. The current carrying loop, thus behaves as a system of two equal and opposite magnetic poles and hence is a magnetic dipole.
The magnetic dipole moment of the current loop (M) is directly proportional to (i) strength of current (i) through the loop and (ii) area (A) enclosed by the loop.
ie, $M\propto I$ and $M\propto A$
$\therefore M=kIA$ ....(i)
where k is constant of proportionality.
If we define unit magnetic dipole moment as that of a small one turn loop of unit area carrying unit carrying unit area current, then from Eq. (i)
$1=k\times 1\times 1$ or $k=1$
$\therefore$ From Eq (i) M = IA
For N such turn M = NIA
Now, length of given wire $L=2\pi r$
or $r=\frac{L}{2\pi }$
Now, area of the coil, $A=\pi r^2=\frac{\pi L^2}{4{\pi }^2}$
$=\frac{L^2}{4\pi }$
Hence, magnitude of magnetic dipole moment is
$M=IA=\frac{I{L}^2}{4\pi }$