Magnets — Question 10

Question 10

Translation and rotation give the same total work along two paths

A permanent dipole m is moved quasistatically from field magnitude B1 to B2. The local field direction is +x throughout the constrained path. Initially the moment is parallel; finally it is antiparallel. Neglect dissipation.

1. Compute total external work by first translating while parallel and then rotating at B2.

2. Compute it by first rotating at B1 and then translating while antiparallel. Compare the totals.

Question 10 - Solution

Translation and rotation give the same total work along two paths

(a) Method and calculation

$$W_{\mathrm{translate,A}}=-m(B_2-B_1),\quad W_{\mathrm{rotate,A}}=2mB_2.$$$$\boxed{W_A=m(B_1+B_2)}.$$

(b) Interpretation and further calculation

$$W_{\mathrm{rotate,B}}=2mB_1,\quad W_{\mathrm{translate,B}}=m(B_2-B_1).$$$$\boxed{W_B=m(B_1+B_2)=U_f-U_i=W_A}.$$

Physical check

Individual work contributions depend on the path, but conservative quasistatic total work depends only on the endpoints.