Explanation

(a) 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.
[image:5_image47.emf]
ie, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M \propto i$ and $M \propto A$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = kiA\ \ \ \ \ \ \ \ \ \ \ldots(i)$
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 current, then from Eq.(i)
$1 = k \times 1 \times 1$ or $k = 1$
$\therefore\ $From Eq.(i)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = iA$
For N such turns
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = NiA$
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
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = iA = \frac{iL^{2}}{4\pi}$