91Ó°ÊÓ

1. The radius of the inner loop = $r_1=20.0cm=0.20m$
2. The radius of the outer loop = $r_2=30.0cm=0.30m$
3. The current through each loop =$i=7.00A$

The magnetic dipole moment of a current carrying loop depends on the number of loops, the current through the loop, and the area enclosed by each turn of the loop. Substitute these values in the formula, and find the magnitude of the magnetic dipole moment. The direction of the magnetic dipole moment is given by the right-hand rule.

$\mathit{μ}\mathbf{=}\mathit{N}\mathit{i}\mathit{A}$

The magnetic moment is given as

$$\begin{gathered} \mathrm{μ}=NiA \\ N=1andA=\mathrm{π}{r}^2 \end{gathered}$$

Since the current in both the loops is flowing in the same direction, the magnetic dipole moment is also in the same direction. Hence, the net moment will be the addition of the two moments.

$$\begin{gathered} ∴\mathrm{μ}={\mathrm{μ}}_1+{\mathrm{μ}}_2 \\ \mathrm{μ}=i\mathrm{π}{r}_1^2+i\mathrm{π}{r}_2^2 \\ \mathrm{μ}=i\mathrm{π}\left(r_1^2+r_2^2\right) \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \mathrm{μ}=7.00×3.14×\left(0.{20}^2+0.{30}^2\right) \\ \mathrm{μ}=7.00×3.14×\left(0.04+0.09\right) \\ \mathrm{μ}=7.00×3.14×\left(0.13\right) \\ \mathrm{μ}=2.86A.m^2 \end{gathered}$$

When the current in the inner loop is reversed, the direction of its magnetic moment will also be reversed. Hence the net magnetic moment will be

$$\begin{gathered} \mathrm{μ}={\mathrm{μ}}_2-{\mathrm{μ}}_1 \\ \mathrm{μ}=i\mathrm{π}{r}_2^2-i\mathrm{π}{r}_1^2 \\ \mathrm{μ}=i\mathrm{π}\left(r_2^2-r_1^2\right) \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \mathrm{μ}=7.00×3.14×\left(0.{30}^2-0.{20}^2\right) \\ \mathrm{μ}=7.00×3.14×\left(0.09-0.04\right) \\ \mathrm{μ}=7.00×3.14×\left(0.05\right) \\ \mathrm{μ}=1.10A.m^2 \end{gathered}$$