Torque is given by $$\vec{\tau}=\vec{r}\times\vec{F}$$ and it is a vector quantity. This means the torque has to be perpendicular to both the force and the radius. However, this doesn't seem to be the case for the question below:

enter image description here

Where the answer says the torque goes into the page enter image description here

Doesn't this contradict the definition of torque given above?


4 Answers 4


While the torque equation and answer is correct, the statement you gave is wrong, I.e., torque is not necessarily perpendicular to the radius.

It is perpendicular to the displacement vector $r$, between point of application and centre. Here force on wire AD is downwards and BC upwards. So it constitutes a couple force, where there is net force but there is net torque.

The wire rotates clockwise as seen from front. You may want to refer to how a electric motor works and the coil rotates.

Hope it helps. enter image description here

  • $\begingroup$ I should have clarified, by radius I meant exactly the vector pointing from the application point to the centre. Still, the answer should still not be correct, as "into the page" is not perpendicular to force vector also required by the equation, which also points into the page. $\endgroup$ Jun 1, 2022 at 11:23
  • $\begingroup$ @joshuamason, look you are directly applying the formula and hence getting it wrong. Instead, draw the F vector on wires AD and BC, then calculate torque about the centre. You will understand why the coil rotates. $\endgroup$
    – PSR_123
    Jun 1, 2022 at 13:07
  • $\begingroup$ And btw, “Torque is into the plane” is given with respect to wire BC only. The same torque applies on AD outwards and hence rotation occurs. If you are still not clear, should I add a pic regarding this in the answer ? $\endgroup$
    – PSR_123
    Jun 1, 2022 at 13:09
  • $\begingroup$ I think I might have misunderstood the meaning of torque. Is it a vector? Or something else? Can you define the notion of torque in your answer? And yes, a pic would help. $\endgroup$ Jun 1, 2022 at 23:32

The question is flawed in that the best you can do is to find the magnitude $(1.68\rm\,N)$ and direction (into the screen) of the force on side $BC$.

What is not given in the question is the axis about which you need to find the torque.

Let us assume it is the vertical axis which bisects the coil about which the torque is to be found.
Then your $\vec r$ has a magnitude of $0.03\,\rm m$ and is directed to the right.

So the direction of the torque is

$\tau = \vec r \times \vec F \Rightarrow \hat {\rm right} \times \hat{\text {into screen}} = \hat {up}$ ie anticlockwise when look up from the bottom.

  • $\begingroup$ I understand everything until the last line. How do we get an almost vector-field like direction (that is, anticlockwise) from a vector calculation? $\endgroup$ Jun 1, 2022 at 11:24

Of the torque definition can answer the question.

Each infinitesimal segment of wire with length ${\rm d}s$, direction $\vec{e}$ and located at $\vec{r}$ has an infinitesimal force ${\rm d}\vec{F}$ applied to it.

The net force applied is

$$\vec{F}_{\rm net} = \int {\rm d} \vec{F}$$

The net torque on the wire is adding up all the little segments.

$$\vec{\tau}_{\rm net} = \int \vec{r} \times \,{\rm d}\vec{F}$$

and due to magnetism the force is defined for each segment as ${\rm d}\vec{F} = (\vec{i} \times \vec{B}) {\rm d}s$

Now to answer your question, you need to evaluate the integral

$$ \vec{\tau}_{\rm net} = \int_0^\ell \vec{r} \times (\vec{i} \times \vec{B}) {\rm d}s$$

where the location varies along the wire $\vec{r} = \vec{r}(s)$.

Note that if the end-points of the wire are designated at $\vec{r}_1$ and $\vec{r}_2$ and the wire is a straight line you have

$$ \vec{r}(s) = \left(1 - \tfrac{s}{\ell}\right) \vec{r}_1 + \left(\tfrac{s}{\ell}\right) \vec{r}_2 $$

with the net torque

$$ \vec{\tau}_{\rm net} = \left( \int_0^\ell \vec{r} {\rm d}s \right) \times (\vec{i} \times \vec{B}) = \frac{ \vec{r}_1 + \vec{r}_2}{2} \times (\vec{i} \times \vec{B})$$

which is interpreted as the force acting on the center of mass of the wire, and then using $$\tau_{\rm net} = \vec{r}_{\rm com} \times \vec{F}_{\rm net}$$


In a rotating object, the velocity and acceleration vectors are different for each point and are continuously changing direction. To deal with this, the rotational quanties are defined as vectors directed along the axis of rotation which is usually not changing direction. This includes the torque vector (which you normally want to be in the same direction as the angular acceleration vector). The direction of the torque vector as indicated by a cross product is consistent with this convention. In this problem, the force on the right side of the loop is into the page, but the torque (relative to the axis of rotation) as a vector is up along the axis.

  • $\begingroup$ yup, that was my train of thought as well. But the answer says otherwise. Can you explain why and how the answer did what it did? $\endgroup$ Jun 1, 2022 at 11:26
  • $\begingroup$ As others have mentioned, the given answer refers to the direction of the force on the right side of the loop, and this is technically not correct for the direction of the torque as a vector. $\endgroup$
    – R.W. Bird
    Jun 1, 2022 at 13:26
  • $\begingroup$ I don't understand what you mean $\endgroup$ Jun 1, 2022 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.