Question

A conducting wire carrying a steady current \(I\) is shaped as shown in the figure below. All connected straight segments meet at right angles. What is the magnetic moment of the current loop?

• A. \(I a b (\hat{j} + \hat{k})\)
• B. \(I a b (\hat{j} - \hat{k})\)
• C. \(\sqrt{2} I a b (\hat{j} + \hat{k})\)
• D. \(I a b (\hat{k} - \hat{j})\)

Hint:

To find the area vectors of non-planar loops quickly, project the loop onto the principal planes (\(xy\), \(yz\), \(xz\)).
The magnetic moment vector is simply the sum of the moments of these projected 2D loops.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question asks for the net magnetic dipole moment of a non-planar loop of wire carrying a steady current \(I\).
The loop consists of six straight segments oriented along the coordinate axes.

Step 2: Key Formula or Approach:

• The magnetic moment of a closed loop of area vector \(\vec{A}\) carrying current \(I\) is:
\[ \vec{M} = I \vec{A} \]
• A complex non-planar loop can be analyzed by decomposing it into multiple planar closed loops.

• We do this by adding and subtracting fictitious currents along a shared edge.

Step 3: Detailed Explanation:

• Let us add a virtual current \(I\) along the line segment from \((-b, 0, 0)\) to \((0, 0, 0)\) and an equal and opposite current from \((0, 0, 0)\) to \((-b, 0, 0)\).

• This decomposes our single 3D loop into two independent closed 2D loops:

• Loop 1 (Vertical, in the \(x-y\) plane):
The current flows in a rectangle of dimensions \(a \times b\).
Tracing the path: \((0,0,0) \to (0,a,0) \to (-b,a,0) \to (-b,0,0) \to (0,0,0)\).
The normal to this loop is in the \(+\hat{k}\) direction.
\[ \vec{M}_1 = I (a b) \hat{k} \]
• Loop 2 (Horizontal, in the \(x-z\) plane):
The current flows in a rectangle of dimensions \(b \times a\).
Tracing the path: \((-b,0,0) \to (-b,0,a) \to (0,0,a) \to (0,0,0) \to (-b,0,0)\).
The normal to this loop is in the \(-\hat{j}\) direction.
\[ \vec{M}_2 = I (a b) (-\hat{j}) \]
• The net magnetic moment is the vector sum of these two components:
\[ \vec{M} = \vec{M}_1 + \vec{M}_2 = I a b \hat{k} - I a b \hat{j} = I a b (\hat{k} - \hat{j}) \]

Step 4: Final Answer:
The magnetic moment of the current loop is \(I a b (\hat{k} - \hat{j})\).