I have a very silly doubt, and it's bothering me. Magnetic Force because of a line current is:

$$\mathbf{F_{mag}} = I\int\mathrm{d}\mathbf{l\times B}$$

However, in a constant magnetic field, for closed circular loops, force is zero as limits of integration goes from $0$ to $0$. (From one refrence point 0 to back at it again)

enter image description here

My doubt is, the limits of integration for a closed circular loop of radius $R$ should go from $0$ to $2\pi R$. Then how is $\mathrm{d}\mathbf{l}$ integrated over the entire loop zero?!

  • 1
    $\begingroup$ Try writing the integral for the loop. Why do you think the limits is 0 to 0? $\endgroup$ Commented Dec 10, 2020 at 7:25

3 Answers 3


It depends which integral you calculate, the line integral of a scalar field or the line integral of a vector field. In books usually $ds$ means scalar field and $d\textbf{s}$ means vector field. I will give you example for both for a circular loop.

First you need a parametrization of a circle with radius R in 3D, for example: $$\gamma(s)=(Rcos(s), Rsin(s), 0)$$

Scalar field

$$\int_{S} ds = \int_{0}^{2\pi}|\dot{\gamma}| ds = \int_{0}^{2\pi}R ds = 2\pi R$$

Vector field

Let $\vec{B}$ be some constant vector field then: $$\int_{S} \vec{B} \cdot d\textbf{s} = \int_{0}^{2\pi}\vec{B} \cdot \dot{\gamma} ds = R\vec{B} \cdot \int_{0}^{2\pi}(-sin(s), cos(s), 0) ds = R\vec{B} \cdot (0,0,0) = 0$$ Therefore we indeed get that the line integral of a constant vector field over a circle with radius R is zero.


For the simple case of a circular loop.

enter image description here

$\displaystyle F= \int_0^\theta B\,I\, Rd\theta \, sin\theta$ and try it for $\theta = \pi$ and $\theta = 2\pi$.

Note the symmetry of the system in that an element on the left has an equal magnitude force and opposite in direction to an element on the right producing a net zero force.


I think you have mis-interpreted what is meant by $d\overrightarrow l$ and $dl$
What we use in calculating $F_{mag}$ is $I \int{d\overrightarrow l \times \overrightarrow B}$ and not $I\int {dl B}$
I too had a tough time figuring out the difference between two once in the beginning.
Actually $dl$ is a scalar quantity. It simply means an infinitesimally small unit of length. When we integrate $dl$
We have, $\int dl = \Delta l$
where l is the total length of the wire, rope or any such body.
But, $d\overrightarrow l =dx\hat i + dy\hat j + dz\hat k$
And integrating it would lead to $\Delta x\hat i + \Delta y\hat j + \Delta z\hat k $, which is the difference in the position vector of initial and final position.

For a closed loop
$\int _{\overrightarrow{r_i}} ^{^\overrightarrow{r_f}}d\overrightarrow l = \overrightarrow{r_f} - \overrightarrow{r_i}$
and by what I guess a closed loop is $\overrightarrow{r_i}=\overrightarrow{r_f}$, which leaves us with
$\int _{\overrightarrow{r_i}} ^{^\overrightarrow{r_f}}d\overrightarrow l = \overrightarrow{r_f} - \overrightarrow{r_i}=0$

  • $\begingroup$ Oh okay! I shouldn't take my notations so lightly! Thank you for the clarification!! $\endgroup$
    – Ruchi
    Commented Dec 10, 2020 at 8:14

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