Question

What is the magnitude of the magnetic dipole moment $\overrightarrow{\mathbf{\mu }}$ of the solenoid described in Problem 51?

Short answer

The magnitude of the magnetic dipole moment is$0.46\text{A}\cdot {\text{m}}^{\text{2}}$.

Step-by-step solution

Listing the given quantities:

The length,$l=25\text{\hspace{0.17em}\hspace{0.17em}cm}=0.25\text{m}$

The diameter,$d=10\text{\hspace{0.17em}\hspace{0.17em}cm}=0.10\text{m}$

The number of turns,$N=200$

The current,$i=0.29\text{\hspace{0.17em}A}$

Understanding the concept of magnetic field:

The magnitude of the magnetic dipole moment can be defined as the product of number of turns, current in that loop, and the area of that loop. Using this concept, you can calculate the magnitude of the magnetic dipole moment.

Formula:

The magnetic dipole moment is,

$\mu =NiA$

Here, $N$is the number of turns,$i$ is the current,$A$ is the area.

Calculations of the magnitude of the magnetic dipole moment:

To calculate the radius of the loop as

$\begin{array}{rcl}R& =& \frac{d}{2}\\ & =& \frac{0.10\text{m}}{2}\\ & =& 0.05\text{m}\end{array}$

Now the magnetic dipole moment is given by,

$\begin{array}{rcl}\mu & =& NiA\\ & =& Ni\pi R^2\\ & =& 200\times 0.29\text{A}\times 3.14\times {\left(0.05\text{m}\right)}^2\\ & =& 0.46\text{A}\cdot {\text{m}}^2\end{array}$

Hence, the magnitude of the magnetic dipole moment is $0.46\text{A}\cdot {\text{m}}^2$.