Fig. 28-49 shows a current loop ABCDEFAcarrying a current i= 5.00 A. The sides of the loop are parallel to the coordinate axes shown, with AB= 20.0 cm, BC= 30.0 cm, and FA= 10.0 cm. In unit-vector notation, what is the magnetic dipole moment of this loop? (Hint:Imagine equal and opposite currents iin the line segment AD; then treat the two rectangular loops ABCDA and ADEFA.)

The current through the loop = i = 5.0 A

The sides of the loop are parallel to the coordinate axes as shown in the figure.

The length of the sides, AB = 20.0 cm = 0.20 m, BC = 30.0 cm = 0.30 m and FA = 10.0 cm = 0.10 m

The current flowing through the loop ABCDEFAwill produce a magnetic dipole moment. But to determine its value, we need to breakdown the loop into two smaller loops, ABCDA and ADEFA. We can calculate their magnetic dipole moments independently and perform the vector addition of the two.

Formula:

$\mu =NiA$

The loop ABCDEFA is broken down into two smaller loops namely,ABCDA and ADEFA.

We imagine equal and opposite currentsi in the imaginary segment AD.

The magnetic moment of the loop ABCDA is calculated as

$$\begin{gathered} \mu =NiA \\ N=1andA=Areaofrec\tan gleABCD \\ {\mu }_1=i\times \left(AB\times BC\right) \\ {\mu }_1=5.0\times 0.20\times 0.30 \\ {\mu }_1=0.30A·m^2 \end{gathered}$$

The direction of the magnetic moment is given by the right-hand rule. By applying the rule, we find that the direction is along -Z axis since the current is flowing clockwise.

${\mu }_1=-0.30\hat{k}A·m^2$

Now the magnetic moment of the loop ADEFA is calculated as

$\mu =NiA$

Here, N = 1 and A = Area of rectangle ADEF

${\mu }_2=i\times \left(AD\times FA\right)$

Since AD = BC = 0.30 m ,

$$\begin{gathered} {\mu }_2=5.0\times 0.30\times 0.10 \\ {\mu }_2=0.15A·m^2 \end{gathered}$$

The direction of the magnetic moment is given by the right-hand rule. By applying the rule, we find that the direction is along +Y axis since the current is flowing clockwise and the loop is parallel to the Z axis.

${\mu }_2=+0.15\hat{j}A·m^2$

The net magnetic moment of the complete loop is

$$\begin{gathered} \stackrel{\rightharpoonup }{\mu }=\stackrel{\rightharpoonup }{{\mu }_1}+\stackrel{\rightharpoonup }{{\mu }_2} \\ \stackrel{\rightharpoonup }{\mu }=-0.30\hat{k}++0.15\hat{j} \end{gathered}$$

Thus,

$\stackrel{\rightharpoonup }{\mu }=\left(0.15\hat{j}-0.30\hat{k}\right)A·m^2$