A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are

Q: Why can we just add fictitious currents?

Because adding equal and opposite currents on the same segment results in zero net change to the physical system, but mathematically allows us to decompose the loop into simpler parts.

Q: How do I know which way the area vector points?

Curl the fingers of your right hand in the direction of the current flow around the loop; your thumb points in the direction of the area vector.

Q: Why is the area vector of ADEFA along the Y-axis?

The loop lies in the XZ plane. A vector perpendicular to the XZ plane must be along the Y-axis. The right-hand rule confirms it's the positive Y-axis.

Q: What if the side lengths were different for the two loops?

You would just calculate the area for each loop separately and use those in your vector addition.

Q: Can this trick be used for any 3D loop?

Yes, any complex 3D loop can be decomposed into a set of planar loops by adding and subtracting currents along shared boundaries.

Question

A wire carrying current $I$ is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$, then the magnitude and direction of magnetic moment of the loop ABCDEFA is

Options:

(A) $abI$, along $\left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$

(B) $\sqrt{2}abI$, along $\left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$

(C) $\sqrt{2}abI$, along $\left(\frac{\hat{j}}{\sqrt{5}} + \frac{2\hat{k}}{\sqrt{5}}\right)$

(D) $abI$, along $\left(\frac{\hat{j}}{\sqrt{5}} + \frac{2\hat{k}}{\sqrt{5}}\right)$

Sigma Percentile · Accepted Answer

### Analyzing the 3D Loop

When you first look at a complex 3D current loop like ABCDEFA, finding its magnetic moment directly might seem a bit intimidating because it doesn't lie in a single plane. But don't worry, physics often provides elegant tricks to simplify such problems.

We can imagine this complex loop as a combination of two simple 2D rectangular loops. How do we do that? By invoking the **Principle of Superposition**. We add a fictitious current $I$ from A to D, and another fictitious current $I$ from D to A. Since they are equal and opposite, they cancel each other out perfectly, meaning we haven't changed the original physical system at all! However, mathematically, this allows us to split the 3D loop into two separate, easy-to-analyze 2D loops.

### Magnetic Moment of Loop ABCDA

First, let's focus on the rectangle ABCDA lying flat in the XY plane. The current flows from A to B, then to C, to D, and back to A.

Using the right-hand rule, if we curl our fingers along this current loop, our thumb points straight up along the positive Z-axis. So, its magnetic moment, $\mathbf{m}_1$, is simply the current $I$ times the area $ab$, in the $\hat{k}$ direction.

$$\mathbf{m}_1 = I(ab)\hat{k}$$

### Magnetic Moment of Loop ADEFA

Now, let's bring back the second rectangle, ADEFA, which stands vertically in the XZ plane. Here, the current flows from A to D, to E, to F, and back to A.

Applying the right-hand rule again for this vertical loop, our thumb points along the positive Y-axis. Thus, its magnetic moment, $\mathbf{m}_2$, is $I$ times $ab$, in the $\hat{j}$ direction.

$$\mathbf{m}_2 = I(ab)\hat{j}$$

### Total Magnetic Moment

The total magnetic moment of our original 3D loop is simply the vector sum of these two individual moments. So, we add $\mathbf{m}_1$ and $\mathbf{m}_2$ together.

$$\mathbf{m} = \mathbf{m}_1 + \mathbf{m}_2 = Iab\hat{k} + Iab\hat{j} = Iab(\hat{j} + \hat{k})$$

Finally, let's find the magnitude and direction. The magnitude is the square root of the sum of squares:

$$|\mathbf{m}| = \sqrt{(Iab)^2 + (Iab)^2} = \sqrt{2}Iab$$

The direction is the vector divided by its magnitude:

$$\hat{m} = \frac{\mathbf{m}}{|\mathbf{m}|} = \frac{Iab(\hat{j} + \hat{k})}{\sqrt{2}Iab} = \frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}$$

This perfectly matches our option (b)!

Select Answer:

Visualized Solution

Analyzing the 3D Loop

Principle of Superposition

Magnetic Moment of Loop ABCDA

Calculating $m_{1}$

Magnetic Moment of Loop ADEFA

Calculating $m_{2}$

Total Magnetic Moment

Magnitude and Direction

The Sigma Insight: Magnetic Moment of Current Loop

Analyzing the 3D Loop

Magnetic Moment of Loop ABCDA

Magnetic Moment of Loop ADEFA

Total Magnetic Moment

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Select Answer:

Visualized Solution

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Principle of Superposition

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Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​

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