Question

What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5.0 cm at a distance of 50 cm from its mid-point? The magnetic moment of the bar magnet is 0.40 Am²

Answer 1

Step 1:Given:

Distance, $r=50\text{cm}$.

Magnetic moment, $m=0.40{\text{Am}}^2$.

Step 2:To Find:

The magnitude of the equatorial and axial fields.

Step 3:Calculate magnitude of the equatorial fields $\left(B_{eq}\right)$:

$$\begin{gathered} B_{eq}=\frac{{\mu }_{0}m}{4\pi r^3} \\ =\frac{\left({10}^{-7}\right)\left(0.40\right)}{4\pi {\left(0.5\right)}^3} \\ =3.3\times {10}^{-7}\text{T} \end{gathered}$$

Here, ${\mu }_0$ is the constant of permeability.

Step 4:Calculate magnitude of the axial fields $\left(B_{axial}\right)$:

$$\begin{gathered} B_{axial}=\frac{2{\mu }_{0}m}{4\pi r^3} \\ =\frac{2\left({10}^{-7}\right)\left(0.40\right)}{4\pi {\left(.5\right)}^3} \\ =6.4\times {10}^{-7}\text{T} \end{gathered}$$

Hence, magnitude of the equatorial fields is $3.3\times {10}^{-7}\text{T}$ ,and the axial field is $6.4\times {10}^{-7}\text{T}$.