Question

What is the magnitude of magnetic dipole moment for a wire of length of $L$ (in m) carrying current $I$ (in Ampere) that is bent in the form of a circle?
A. ${\displaystyle \frac{I{L}^2}{4\pi }}$
B. ${\displaystyle \frac{I^2{L}^2}{4\pi }}$
C. ${\displaystyle \frac{I^2{L}^{}}{8\pi }}$
D. ${\displaystyle \frac{I{L}^2}{8\pi }}$

Answer 1

Hint
Magnetic dipole moment is generally calculated for electric current loops. It is given as the product of the area encapsulated by the loop and the total electric current flowing through the loop.
$\Rightarrow M=I\times A$
where $M$ is the magnetic dipole moment, $I$ is the current and $A$ is the area encompassed by the loop.

Complete step by step answer
The magnetic dipole moment is used to represent the magnetic strength of any object that produces a magnetic field. In the given question, we are asked to find the dipole moment for a wire with the following general properties:
Length of the wire = $L$
Current through the wire = $I$
Since the wire is bent like a circle, it forms a loop of radius R with the area given as:
$\Rightarrow A=\pi R^2$
As $R$ is not explicitly specified in the question, we need to substitute it with the given quantity (i.e. $L$ ). For this, we know that the circumference of a circle is given by:
$\Rightarrow C=2\pi R$
This value should be equal to the length of the wire.
Hence,
$\Rightarrow L=2\pi R$
$\Rightarrow R={\displaystyle \frac{L}{2\pi }}$ [Solving for R] [Eq. 1]
Putting these values in the formula of magnetic dipole moment, we get:
$\Rightarrow M=I\times A$
$\Rightarrow M=I\times \pi R^2$
Substituting the value of $R$ from [Eq. 1] gives us:
$\Rightarrow M=I\times \pi {\left({\displaystyle \frac{L}{2\pi }}\right)}^2$
$\Rightarrow M={\displaystyle \frac{I\times \pi \times L^2}{4{\pi }^2}}$
After cancelling and multiplication, we get:
$\Rightarrow M={\displaystyle \frac{I{L}^2}{4\pi }}$
$\therefore$ The answer is option (A).

Note
From this exercise, we learn that the magnetic dipole moment for a generic loop is given by $M={\displaystyle \frac{I{L}^2}{4\pi }}$ . The result can directly be used for any other similar problems. The magnetic moment is a vector quantity. Hence, the right hand rule should be used to determine the direction of the moment.