State that a current carrying loop behaves as a magnetic dipole. Hence write an expression for its magnetic dipole moment .

Explanation

The magnetic field from the centre of a circular loop of radius R along the axis is given by
`vecB = (mu_(0)I)/ 2 R^(2)/((R^(2) + z^(2))^(3/2))hatk`
At large distance `z gt gt R`, therefore `R^(2) + z^(2) approx z^(2)`, we have
`vecB = (mu_(0)I)/2 R^(2)/z^(3) hatk` .....(1)
Let A be the area of the circular loop `A = pi R^(2)`.
So rewriting the equation (1) in terms of area of the loop , we have
`vecB = (mu_(0)I)/ (2 pi) A/z^(3) hatk`
` vecB = (mu_(0))/(4 pi) (2 I A)/z^(3) hatk ` ...(2)
Comparing equation (2) with equation (1) dimensionally, we get
`p_(m) = IA` ltbr. where `p_(m) ` is called magnetic dipole moment . In vector notation,
`vecp_(m) = I vecA` ......(3)
This implies that a current carrying circular loop behaves as a magnetic dipole of magnetic moment `vecp_(m)` . So, the magnetic dipole moment of any current loop is equal to the product of the current and area of the loop.