Force components on dipole falling through copper ring

Question

We have a magentic dipole parallel with z-axis (shown on the diagram below) falling through copper loop, because of changing magnetic flux there is induced current in cooper ring/loop and consequently we get magnetic force on ring/loop and on our dipole (because of Newton's thrid law).

Lets start calculating force on our dipole with $$F_{dipole} = - \int dF_{ring} = -\int I(\vec{dl} \times \vec{B})$$ we know that $\vec{dl} = R d\phi \hat{\phi}$ and $\vec{B} = B_z \hat{z} + B_{\rho}\hat{\rho}$, continuig with integration... $$= -\int_0^{2\pi} IRB_z \; d\phi \; \hat{\rho} +\int_0^{2\pi} IRB_{\rho} \; d\phi \; \hat{z} = -F_{\rho} \hat{\rho} + F_z \hat{z}$$

We get that there is a forcce $F_{\rho}$ that tries to compress the dipole in all directions, but my intuition tells me that force in radial direction should be zero ($F_{\rho} = 0$). Where did I go wrong? Plese explain, thank you very much in advance.

Answer 1

Your error is assuming that $$ \int_0^{2\pi} \left[ IRB_z \hat{\rho}\right] \; d\phi = \left[\int_0^{2\pi} IRB_z\; d\phi \right]\hat{\rho}. $$ This is because at different points on the wire, the vector $\hat{\rho}$ points in different directions. It is therefore not a constant vector and cannot be factored out of the integral.

If you want to do this integral correctly, you can rewrite $\hat{\rho}$ in terms of $\hat{x}$ and $\hat{y}$, which are constant unit vectors at everywhere in space: $$ \int_0^{2\pi} \left[ IRB_z \hat{\rho}\right] d\phi = \int_0^{2\pi} \left[ IRB_z (\cos \phi \,\hat{x} + \sin \phi \,\hat{y} ) \right] d\phi \\= IRB_z \left\{ \left[ \int_0^{2\pi} \cos \phi \, d \phi \right] \hat{x} + \left[ \int_0^{2\pi} \sin \phi \, d \phi \right] \hat{y} \right\} $$ If you do both of these integrals, though, you get zero. So all this tells you is that there are no horizontal components to the net force on the force on the dipole.