Irodov 3.243: Weighing a Loop by Its Centre – Amit & Deepali Sinha Foundation

Problem Irodov 3.243

A thin circular loop of radius $R = 100$ mm carries a current, and the magnetic induction at its centre is $B = 6.0\ \mu$ T. Find the magnetic moment of the loop.

The thin loop of radius R and its centre O, where the induction is B. — The thin loop of radius $R$ and its centre $O$ , where the induction is $B$ .

Given

$R = 100$ mm
$B = 6.0\ \mu$ T at the centre

Find

the magnetic moment

p_m

Solution

Step 1The current from the field at the centre

Every element $d\vect l$ of the loop is square to the radius joining it to $O$ , so by Biot and Savart each contributes $\mu_0 I\,dl/(4\pi R^2)$ along the axis, all in the same sense. Summing round the circumference $2\pi R$ :

B = \frac{\mu_0 I}{4\pi R^2}\cdot 2\pi R = \frac{\mu_0 I}{2R}

Solving for the current:

I = \frac{2RB}{\mu_0} \tag{1}

Step 2The moment

The loop with its current I drawn counter-clockwise: B at the centre points out of the page, and so does the moment p_m, by the same right-hand rule. — The loop with its current $I$ drawn counter-clockwise: $\vect B$ at the centre points out of the page, and so does the moment $\vect p_m$ , by the same right-hand rule.

The magnetic moment of a plane loop is its current times the area it encloses, $p_m = I\,\pi R^2$ . With (1):

p_m = \frac{2RB}{\mu_0}\,\pi R^2

Numerically,

p_m = \frac{2\pi\,(0.100\text{ m})^3\,(6.0\times 10^{-6}\text{ T})}{4\pi\times 10^{-7}\text{ H/m}}

p_m = \frac{2\pi R^3 B}{\mu_0} = 3.0\times 10^{-2}\ \text{A}\cdot\text{m}^2

(2)

The moment points along the axis, the same way as $\vect B$ at the centre.

Check

Units: $\text{m}^3\cdot\text{T}/(\text{H/m}) = \text{m}^4\,\text{T/H}$ ; with $\text{T} = \text{Wb/m}^2$ and $\text{H} = \text{Wb/A}$ this is $\text{A}\cdot\text{m}^2$ , a moment. The current is $I = 2RB/\mu_0 = 0.95$ A, and $\pi R^2 I = 0.0314\text{ m}^2 \times 0.95\text{ A}$ gives the same $0.030$ A $\cdot$ m $^2$ .

Remark

At a distance $z$ far along the axis the loop’s field is that of a dipole, $\mu_0 p_m/(2\pi z^3)$ . Written with (2), it is $(R/z)^3$ times the given centre value $B$ : the field at the centre sets the whole far axial field, which falls to one thousandth of its centre value about ten radii away.