Question

Two concentric, circular wire loops, of radii r₁ = 20.0cm and r₂ = 30.0 cm, are located in an xyplane; each carries a clockwise current of 7.00 A (Figure).

(a) Find the magnitude of the net magnetic dipole moment of the system.

(b) Repeat for reversed current in the inner loop.

Short answer

1. The magnitude and the net dipole moment is$\mu =2.86A.m^2$
2. The dipole moment for the reversed current in the inner loop is $\mu =1.10A.m^2$

Step-by-step solution

Write the given data:

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 =$i=7.00A$

Determine the formula for the dipole moment as:

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.

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

(a) Calculate the magnitude of the net magnetic dipole moment of the system

The magnetic moment is given as

$$\begin{gathered} \mu =NiA \\ N=1andA=\pi r^2 \end{gathered}$$

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.

$$\begin{gathered} \therefore \mu ={\mu }_1+{\mu }_2 \\ \mu =i\pi r_1^2+i\pi r_2^2 \\ \mu =i\pi \left(r_1^2+r_2^2\right) \end{gathered}$$

Substitute the values and solve as:

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

(b) Calculate the magnitude of the net magnetic dipole moment of the system for the current reversed in it

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

$$\begin{gathered} \mu ={\mu }_2-{\mu }_1 \\ \mu =i\pi r_2^2-i\pi r_1^2 \\ \mu =i\pi \left(r_2^2-r_1^2\right) \end{gathered}$$

Substitute the values and solve as:

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