# Prove that the torque of any current loop is m X B

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1. May 30, 2015

### Happiness

1. The problem statement, all variables and given/known data
Problem 6.2 of Griffith's "Introduction to Electrodynamics": Starting from the Lorentz force law $\vec F=\int I (d\vec l \times \vec B)$, show that the torque on any steady current distribution (not just a square loop) in a uniform field $\vec B$ is $\vec m\times \vec B$. ($\vec m$ is the magnetic moment.)

2. Relevant equations
Let the torque be $\vec N$.

$d\vec N = \vec r\times d\vec F$.

3. The attempt at a solution
Useful identity: $\oint \vec r\times d\vec l = 2\vec a$, where $\vec a$ is the area of the loop and points perpendicularly to its surface.

My question: the solution says that $d\vec r = d\vec l$, which I don't understand. They are clearly pointing in different directions. $d\vec r$ points in the direction from the origin to the point $r$, while $d\vec l$ points in the direction of the wire of the loop, which in general is different from the direction of $d\vec r$.

The solution:

2. May 30, 2015

### Delta²

It is $\vec{r}$ that points from the origin to the point r =(x,y,z) of the loop and $\vec{r'}=\hat{i}(x+dx)+\hat{j}(y+dy)+\hat{k}(z+dz)$ that points from the origin to the point r'=(x+dx,y+dy,z+dz) of the loop. Their difference is the ifinitesimal vector $d\vec{r}=\hat{i}dx+\hat{j}dy+\hat{k}dz$ which is the same as dl.

3. May 30, 2015

### Happiness

Thanks a lot! I've got it now.