Question

What is the magnetic dipole moment of a $\mathbf{3000}$-turns rectangular coil that measures $\mathbf{3}\mathbf{}\mathbf{cm}$by $\mathbf{5}\mathbf{}\mathbf{cm}$ and carries a current of $\mathbf{2}\mathbf{}\mathbf{A}$ ?

Short answer

The magnetic dipole moment of the rectangular coil is$9\mathrm{A}.{\mathrm{m}}^2$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The number of turns is, $\mathrm{N}=3000$.
• The dimensions of a rectangular coil is, $\left(\mathrm{l}\times \mathrm{b}\right)=\left(3\mathrm{cm}\times 5\mathrm{cm}\right)$.
• The current in the wire is, $\mathrm{I}=2\mathrm{A}$.

Significance of magnetic dipole moment

In this question, the magnetic dipole moment of a rectangular coil will be estimated by taking the product of a total number of turns, electric current and cross-sectional area of the coil. The magnetic dipole moment helps to determine the value of the magnetic field.

Determination the magnetic dipole moment of the rectangular coil

The relation of magnetic dipole moment of the rectangular coil is expressed as,

$\begin{array}{rcl}\mathrm{\mu }& =& \mathrm{NIA}\\ & =& \mathrm{NI}\left(\mathrm{l}\times \mathrm{b}\right)\end{array}$

Here, $\mathrm{\mu }$ is the magnetic dipole moment of the rectangular coil and $\mathrm{A}$ is the cross-sectional area of the rectangular coil.

Substitute all the known values in the above equation.

$\begin{array}{rcl}\mathrm{\mu }& =& \left(3000\right)\left(2\mathrm{A}\right)\left(3\mathrm{cm}\times \frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\left(5\mathrm{cm}\times \frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\\ & =& 9\mathrm{A}.{\mathrm{m}}^2\end{array}$

Thus, the magnetic dipole moment of the rectangular coil is $9\mathrm{A}.{\mathrm{m}}^2$.