Q.Do magnetic forces obey Newton's third law. Verify for two current elements $d\vec{l}_1 = dl\,\hat{i}$ located at the origin and $d\vec{l}_2 = dl\,\hat{j}$ located at $(0, R, 0)$. Both carry current $I$.

Newton's third law requires $\vec{F}_{1\text{ due to }2}=-\vec{F}_{2\text{ due to }1}$. Check this for $d\vec{l}_1=dl\,\hat{i}$ at the origin and $d\vec{l}_2=dl\,\hat{j}$ at $(0,R,0)$, both carrying current $I$.

Force on 2 due to 1: field of element 1 at element 2's location is $d\vec{B}_1=\dfrac{\mu_0 I\,dl}{4\pi R^2}\hat{k}$ (since $\hat{i}\times\hat{j}=\hat{k}$), so $d\vec{F}_{2\text{ due to }1}=I\,d\vec{l}_2\times d\vec{B}_1=\dfrac{\mu_0 I^2 dl^2}{4\pi R^2}\hat{i}$ - nonzero.

Force on 1 due to 2: field of element 2 at element 1's location uses $\hat{r}_{21}=-\hat{j}$, so $d\vec{l}_2\times\hat{r}_{21}=dl(\hat{j}\times(-\hat{j}))=0$, giving $d\vec{B}_2=0$ and hence $d\vec{F}_{1\text{ due to }2}=0$. …

Why Newton's third law can fail here

Newton's third law requires that the force of A on B be equal and opposite to the force of B on A. This holds for complete, closed circuits (verified experimentally), but it is not guaranteed for two isolated current elements $d\vec{l}_1$, $d\vec{l}_2$ considered on their own - because the magnetic force each element feels depends on the relative orientation of the two elements through a cross product, and that geometric relationship is not symmetric in general. (The "missing" momentum is actually carried by the electromagnetic field itself - but that is beyond what we need to verify here.)

Setting up the geometry

• Element 1: $d\vec{l}_1 = dl\,\hat{i}$, located at the origin $(0,0,0)$.
• Element 2: $d\vec{l}_2 = dl\,\hat{j}$, located at $(0,R,0)$.

Both carry current $I$. The unit vector from 1 to 2 is $\hat{r}_{12}=\hat{j}$ (distance $R$); the unit vector from 2 to 1 is $\hat{r}_{21}=-\hat{j}$ (same distance $R$).

Force on element 2 due to element 1

Field at element 2's location, produced by element 1 (Biot-Savart):

Force on element 2 in this field:

This is nonzero, pointing along $+\hat{i}$.

Force on element 1 due to element 2

Field at element 1's location, produced by element 2:

since the cross product of any vector with itself (or its negative) is zero.

So the force on element 1:

Comparing the two …

Method: Testing Newton's Third Law Between Two Current Elements

General technique for verifying (or disproving) $\vec F_{12} = -\vec F_{21}$ between two isolated current elements.

Steps

Step 1: Compute the field each element produces at the OTHER element's location, via Biot-Savart

using the unit vector FROM the source element TO the field point. This direction flips when you swap which element is the source — compute it twice, once each way, carefully.

Step 2: Compute each force from the OTHER element's own current in that field

Pair element 2's own $d\vec l_2$ with the field FROM element 1, and separately element 1's own $d\vec l_1$ with the field FROM element 2 — never reuse one cross-product result for both directions.

Step 3: Compare the two force vectors directly …