Question

A current-carrying wireframe is in the shape of digit eight ( $8$). It is carrying current ($i_o$). If the radius of each loop is ( $R_o$), then the net magnetic dipole moment of the figure is

• A. $\left(i_o\pi {R_o}^2\right)\sqrt{2}$
• B. Zero
• C. $i_o\times 2\pi {R_o}^2$
• D. $i_o\times 4\pi R_o$

Answer 1

The correct option is B

Zero

Step 1: Given data

Current flowing in the upper loop of the wire in the clockwise direction$=i_0$

Current flowing in the lower loop of the wire in the anti-clockwise direction$=i_o$

Radius of the loop$=R_o$

Step 2: Assumptions

Magnetic moment of the upper loop$=M_1$

Magnetic moment of the lower loop$=M_2$

Net magnetic moment$=M$

Step 3: Formula used

$M_{cl}=I\times \pi r^2$ …………………..(a)

Where ($M_{cl}$ ) is the magnetic dipole moment of the circular loop

($I$) is the current flowing through the circular loop

($r$) is the radius of the circular loop

Step 4: Calculation of the net magnetic dipole moment

Substituting the given values in equation (a), the magnetic dipole moment of the upper loop becomes

$M_1=i_o\times \pi {R_o}^2$ ………………..(b)

Applying the right-hand thumb rule, it is clear that the direction of the magnetic dipole moment in the upper loop will be “out of the plane”.

Substituting the given values in equation (a), the magnetic dipole moment of the lower loop becomes

$M_2=i_o\times \pi {R_o}^2$ ………………(c)

Applying the right-hand thumb rule, it is clear that the direction of the magnetic dipole moment in the lower loop will be “into of the plane”.

Since the direction of the magnetic dipole moment in the upper loop and the lower loop are “opposite”, therefore the net magnetic dipole moment will be given by

$M=M_1-M_2$ ……………….(d)

Substituting equation (b) and (c) in equation (d), we get

$$\begin{gathered} M=\left(i_o\pi {R_o}^2\right)-\left(i_o\pi {R_o}^2\right) \\ M=0 \end{gathered}$$

Hence, option (B) is correct.