Question

The on-axis magnetic field strength $10\mathrm{cm}$from a small bar magnet is $5.0\mathrm{\mu T}$.

a. What is the bar magnet’s magnetic dipole moment?

b. What is the on-axis field strength 15 cm from the magnet?

Short answer

a. magnetic moment is $25\mathrm{mA}.{\mathrm{m}}^2$

b. Magnetic field strength will be$1.48\mathrm{\mu T}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

magnetic field strength =$5\mathrm{\mu T}$

distance = $10\mathrm{cm}$

We have to find the magnetic moments.

Simplify

The magnetic field in of a bar magnet in terms of magnetic moment is given by,

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\overrightarrow{\mathrm{\mu }}}{2{\mathrm{\pi R}}^3} \\ \overrightarrow{\mathrm{\mu }}=\frac{2{\mathrm{\pi R}}^3\mathrm{B}}{{\mathrm{\mu }}_0} \\ \overrightarrow{\mathrm{\mu }}=\frac{2\mathrm{\pi }\times {(0.1)}^3\times 5\times {10}^{-6}}{4\mathrm{\pi }\times {10}^{-7}}=25\mathrm{mA}.{\mathrm{m}}^2 \end{gathered}$$

Part (b) Step 1: Given information

We have given,

distance = $15\mathrm{cm}$

We have to find the magnetic field strength.

Simplify

The magnetic field will be

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\overrightarrow{\mathrm{\mu }}}{2{\mathrm{\pi R}}^3} \\ \mathrm{B}=\frac{(4\mathrm{\pi }\times {10}^{-7})\times 0.025}{2\mathrm{\pi }\times {(0.15)}^3} \\ \mathrm{B}=1.48\mathrm{\mu T}. \end{gathered}$$