I was taught that in a magnetic field $\vec B$, a current carrying loop with a magnetic moment $\vec M$, the loop experiences a torque given by $\tau=\vec M\times\vec B$. Further, the associated energy is given by (as in the electrical case by integrating $\tau d\theta$) $U(\theta)=-\vec M\cdot \vec B$. Now I had learnt that magnetic field cannot do any work (force and velocity are always perpendicular).

So,how are we associating an energy here? Where does this energy $U(\theta)$ come from? I can't account for it.

  • $\begingroup$ "Now I had learnt that magnetic field cannot do any work (force and velocity are always perpendicular)." - Take a look at the answer here: "The Lorentz force $\textbf{F}=q\textbf{v}\times\textbf{B}$ never does work on the particle with charge $q$. This is not the same thing as saying that the magnetic field never does work." $\endgroup$
    – Hal Hollis
    Oct 9, 2018 at 15:18

1 Answer 1


@Hal Hollis mentioned correctly in the above comment that $B$ do non zero work on systems, like in this question.

The given potential energy expression is misleading in interpretation and it is derived from the work done by $B$ as follows: $$W_B=\int_{\theta_i}^ {\theta_f} \tau d\theta = \int_{\theta_i}^ {\theta_f}MBsin(\theta)d\theta =-MB[cos(\theta_f) -cos(\theta_i)]$$ The general expression for potential energy change will be $$\Delta U=-W_B=MB[cos(\theta_f) -cos(\theta_i)]$$

Setting reference: zero of potential energy as(when torque is maximum) $\theta_f =\pi/2$ and putting $\theta_i =\theta$, we get $$U(\theta)=-MBcos(\theta)=-\vec{M}.\vec{B}$$ So the energy $U(\theta)$ is nothing but the change in total potential energy of the system with respect to position $\pi/2$ and it is not an absolute energy.


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