Question

The magnetic dipole moment of a current loop carrying current $I$ and of area $A$ with $n$ turns is ________.

• A. $\frac{IA}{n}$
• B. $\frac{IA}{n^2}$
• C. $\frac{IA}{2}$
• D. $nIA$
• E. $IA$

Hint:

Unit of magnetic moment is $A \cdot m^2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This question asks for the definition of the magnetic dipole moment (\(\vec{m}\) or \(\vec{\mu}\)) for a coil of wire with multiple turns. The magnetic dipole moment is a measure of the strength and orientation of a magnetic source.
Step 2: Key Formula or Approach
The magnetic dipole moment of a single planar loop of wire is defined as:
\[ \vec{m}_{single} = I\vec{A} \] where:
- \(I\) is the current flowing in the loop.
- \(\vec{A}\) is the area vector of the loop. Its magnitude is the area A, and its direction is perpendicular to the plane of the loop, given by the right-hand rule.
For a coil with \(n\) identical turns, each turn contributes to the total magnetic moment. The total magnetic moment is the sum of the individual moments.
Step 3: Detailed Explanation
If we have a coil with \(n\) turns, and each turn carries the same current \(I\) and has the same area \(A\), then the magnetic moment of each turn is \(I\vec{A}\).
Since all the turns are wound together, their area vectors point in the same direction. Therefore, the total magnetic moment of the coil is the sum of the moments of the individual turns:
\[ \vec{m}_{total} = \sum_{i=1}^n \vec{m}_i = \sum_{i=1}^n I\vec{A} = n(I\vec{A}) \] The magnitude of the total magnetic dipole moment is:
\[ m_{total} = nIA \] This shows that the magnetic moment is directly proportional to the number of turns, the current, and the area of the loop.
Step 4: Final Answer
The magnetic dipole moment is nIA.