Question

Two identical short bar magnets each having a magnetic moment of $10{\mathrm{Am}}^2$, are arranged such that their axial lines are perpendicular to each other and their centres are along the same straight line in a horizontal plane. If the distance between their centres is $0.2\mathrm{m}$, What is the resultant magnetic induction at a point midway between them? $\left({\mu }_0=4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}\right)$.

Answer 1

Step 1: Diagram of the identical magnets

Step 2: Calculation of the magnetic induction

The net magnetic induction can be calculated using the following expression.

$$\begin{gathered} B_{\mathrm{net}}=\sqrt{B_a^2+B_e^2} \\ {B}_{\mathrm{net}}=\sqrt{{\left(\frac{{\mu }_0}{4\pi }\times \frac{2M}{d^2}\right)}^2+{\left(\frac{{\mu }_0}{4\pi }\times \frac{M}{d^2}\right)}^2} \end{gathered}$$

Here, $B_{\mathrm{net}}$ is the total magnetic induction, $B_1$ is the magnetic induction due to the first magnet, $B_2$ is the magnetic induction due to the second magnet, $M$ is the magnetic moment, $d$ is the distance of the midpoint of the magnets, and ${\mu }_0$ is the permeability of the free space.

$$\begin{gathered} B_{\mathrm{net}}=\sqrt{\frac{{\mu }_0}{4\pi d^2}\left[{\left(2M\right)}^2+{\left(M\right)}^2\right]} \\ {B}_{net}=\sqrt{5}\times \frac{{\mu }_0}{4\pi }\times \frac{M}{d^2} \\ {B}_{net}=\sqrt{5}\times \frac{4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}}{4\pi }\times \frac{10{\mathrm{Am}}^2}{{\left(0.1\mathrm{m}\right)}^2} \\ {B}_{net}=\sqrt{5}\times {10}^{-3}\mathrm{Tesla} \end{gathered}$$

Hence, the resultant magnetic induction at a point midway between them is $\sqrt{5}\times {10}^{-3}\mathrm{Tesla}$.