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(D) Both magnets are placed in the magnetic field of one another. The magnetic field $B_1$ produced by magnet $1$ at the position of magnet $2$ (on it

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Fourth power of the distance

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(D) Both magnets are placed in the magnetic field of one another. The magnetic field $B_1$ produced by magnet $1$ at the position of magnet $2$ (on its axial line) is given by:$B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M_1}{r^3}$The potential energy $U$ of a magnetic dipole $M_2$ in an external magnetic field $B_1$ is given by $U = -M_2 B_1 \cos 	heta$. Since the magnets are placed along the same axis with like poles facing each other,the magnetic moments are aligned,so $	heta = 0^\circ$ and $\cos 0^\circ = 1$. Thus:$U = -M_2 B_1 = -M_2 \left( \frac{\mu_0}{4\pi} \cdot \frac{2M_1}{r^3} ight) = -\frac{\mu_0}{4\pi} \cdot \frac{2M_1 M_2}{r^3}$The force $F$ between the magnets is given by the negative gradient of the potential energy:$F = -\frac{dU}{dr} = -\frac{d}{dr} \left( -\frac{\mu_0}{4\pi} \cdot \frac{2M_1 M_2}{r^3} ight)$$F = \frac{\mu_0}{4\pi} \cdot 2M_1 M_2 \cdot \frac{d}{dr} (r^{-3}) = \frac{\mu_0}{4\pi} \cdot 2M_1 M_2 \cdot (-3r^{-4})$$F = -\frac{\mu_0}{4\pi} \cdot \frac{6M_1 M_2}{r^4}$The magnitude of the force is $F = \frac{\mu_0}{4\pi} \cdot \frac{6M_1 M_2}{r^4}$.Therefore,the force varies inversely as the fourth power of the distance,i.e.,$F \propto \frac{1}{r^4}$.

Two short magnets are placed along the same axis with their like poles facing each other. They repel each other with a force which varies inversely as:

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