Question

The center of a bar magnet whose magnetic dipole moment is $\langle 8,0,0\rangle \mathbf{\text{\hspace{0.17em}A}}\mathbf{\cdot }{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$is located at the origin. A second bar magnet whose magnetic dipole moment is $\langle 3,0,0\rangle \mathbf{\text{\hspace{0.17em}A}}\mathbf{\cdot }{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$is located at $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.12}\mathbf{\text{\hspace{0.17em}m}}$. What is the vector force on the second magnet due to the first magnet?

Short answer

The vector force on the second magnet due to the first magnet is$6.94\times {10}^{-2}\text{\hspace{0.17em}N}$

Step-by-step solution

Given Information

The dipole moment is given by $\overrightarrow{{\mu }_1}=〈8,0,0〉\text{\hspace{0.17em}A}\cdot {\text{m}}^{\text{2}}$at location $〈0,0,0〉$and $\overrightarrow{{\mu }_2}=〈3,0,0〉\text{\hspace{0.17em}A}\cdot {\text{m}}^{\text{2}}$at location $〈0.12,0,0〉\text{\hspace{0.17em}m}$$〈0.12,0,0〉\text{\hspace{0.17em}m}$

Force of Attraction and Repulsion

It is known that two magnets should attract or repel each other with a force proportional to $\frac{1}{r^4}$and the attractive or repulsive force exert on a magnetic dipole moment by this magnet is given as:

$F=3{\mu }_2\left(\frac{{\mu }_0}{4\pi }\cdot \frac{2{\mu }_1}{r^4}\right)\cdots \left(1\right)$

Calculate the vector force

Now substitute the value ${\mu }_1=8,{\mu }_2=3$and $r=0.12$in equation (1),

$\begin{array}{c}F=3{\mu }_2\left(\frac{{\mu }_0}{4\pi }\cdot \frac{2{\mu }_1}{r^4}\right)\\ =3\cdot 3\left(\left(1\times {10}^{-7}\right)\left(\frac{2\left(8\right)}{{\left(0.12\right)}^4}\right)\right)\\ =6.94\times {10}^{-2}\text{\hspace{0.17em}N}\end{array}$

Thus, the vector force on the second magnet due to the first magnet is $6.94\times {10}^{-2}\text{\hspace{0.17em}N}$.