Question

The on-axis magnetic field strength 15cm from a small bar magnet is 4.8 μT

a)What is the bar magnet's magnetic dipole moment?

b)What is the on-axis field strength 19cm from the magnet?

Answer 1

Concepts and reason

The concept required to solve this question is the magnetic field on the axis of the dipole.

Initially, write the expression for the magnetic field on the axis of the bar magnet. Then, rearrange the expression of the magnetic field for the magnetic dipole moment.

The magnetic field on the axis of the bar magnet can be calculated using the calculated value of the dipole moment and the distance.

Fundamentals

The expression of the magnetic field on the axis of the bar magnet is,

$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2p}}{{{z^3}}}$

Here, B is the magnetic field, ${\mu _0}$ is the permeability, p is the magnetic dipole moment, and z is the distance.

(a)

The expression of the magnetic field on the axis of the bar magnet is,

$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2p}}{{{z^3}}}$

Rearrange the expression for the dipole moment.

$p = \frac{{4\pi B{z^3}}}{{2{\mu _0}}}$

Substitute $4.8{\rm{ }}\mu {\rm{T}}$ for B, 15 cm for z, and $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ .

$\begin{array}{c}\\p = \frac{{4\pi \left( {4.8{\rm{ }}\mu {\rm{T}}\left( {\frac{{{{10}^{ - 6}}{\rm{ T}}}}{{1{\rm{ }}\mu {\rm{T}}}}} \right)} \right){{\left( {15{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ cm}}}}{{1{\rm{ cm}}}}} \right)} \right)}^3}}}{{2\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)}}\\\\ = 0.081{\rm{ A}} \cdot {{\rm{m}}^2}\\\end{array}$

(b)

The expression of the magnetic field on the axis of the bar magnet is,

$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2p}}{{{z^3}}}$

Substitute $0.081{\rm{ A}} \cdot {{\rm{m}}^2}$ for p, 19 cm for z, and $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ .

$\begin{array}{c}\\B = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)}}{{4\pi }}\frac{{2\left( {0.081{\rm{ A}} \cdot {{\rm{m}}^2}} \right)}}{{{{\left( {19{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ cm}}}}{{1{\rm{ cm}}}}} \right)} \right)}^3}}}\\\\ = 2.36 \times {10^{ - 6}}{\rm{ T}}\\\\{\rm{ = 2}}{\rm{.36 }}\mu {\rm{T}}\\\end{array}$

Ans: Part a

The magnetic dipole moment is $0.081{\rm{ A}} \cdot {{\rm{m}}^2}$ .

Part b

The magnetic field is ${\rm{ 2}}{\rm{.36 }}\mu {\rm{T}}$ .