A Current-Carrying Coil as a Magnetic Dipole — Question 8

Question 8

Two small coaxial coils attract with an inverse-fourth-power force

Two small coils have fixed moments $\mu_1=.0200$ and $\mu_2=.0300$ ampere square metre, both along +z. Coil 1 is fixed at z=0; coil 2 is on axis at d=.100 m. Both coil radii are much smaller than d. Use $B_1(z)=\mu_0\mu_1/(2\pi z^3)$ and the mechanical interaction potential $U=-\mu_2B_1$. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the axial force on coil 2 and its direction.

2. Find the quasistatic external mechanical work to move coil 2 from d to 2d while both moments remain fixed.

Question 8 - Solution

Two small coaxial coils attract with an inverse-fourth-power force

(a) Method and calculation

$$U(z)=-\frac{\mu_0\mu_1\mu_2}{2\pi z^3},\quad F_z=-\frac{dU}{dz}=-\frac{3\mu_0\mu_1\mu_2}{2\pi z^4}.$$$$\boxed{F_z(d)=-3.60\,\mu\mathrm N}\quad\text{toward coil 1}.$$

(b) Interpretation and further calculation

$$W_{\mathrm{ext}}=U(2d)-U(d)=\frac{\mu_0\mu_1\mu_2}{2\pi d^3}\left(1-\frac18\right).$$$$\boxed{W_{\mathrm{ext}}=1.05\times10^{-7}\,\mathrm J}.$$

Physical check

This is mechanical work at fixed moments; maintaining real coil currents during motion also involves their supplies. The force law assumes the dipole regime throughout.