# How is torque proportional to current and magnetic flux in a DC motor?

It is well known in physics that a rigid body having a current $$\i\$$ in a closed loop under a magnetic field $$\\vec{B}\$$ experiences a torque which is

$$\\vec{T}= \vec{m}\times\vec{B}\$$

where the magnetic moment $$\\vec{m}\$$ is defined as a vector perpendicular to the plain of the coil itself and it has a module equal to the product of the current with its enclosed area.

$$\ \vec{m} = i\vec{S}\$$

My problem is that I would like to understand in which way in a DC motor the torque $$\\vec{T}\$$ on its rotor is proportional to the product of current $$\i\$$ and the magnetic flux $$\\phi(\vec{B})\$$ .

$$\ T = K \phi(\vec{B})i \$$

as reported (without demonstrations!) in an engineering book I am dealing with.

Particularly my need of clarification comes from the following consideration: when the flux $$\\phi(\vec{B})\$$ is zero then the torque $$\\vec{T}\$$ is at its maximum, while, when the flux $$\\phi(\vec{B})\$$ is at its maximum then the torque $$\\vec{T}\$$ is zero.

For the purpose of a clear understanding I report in the following picture a simple scheme of a DC motor: According to Lorentz equation, the force (green arrows) applied to the rotor

$$\ \vec{F} = i\vec{l}\times \vec{B}\$$

which means the force is constant in both module ($$\F = ilB\$$) and direction all over the rotor movement around its axis. What is not constant is the torque $$\\vec{T} = \vec{b}\times\vec{F}\$$, the module of which being

$$\T = ilbB\sin{\alpha}\ = iSB\sin{\alpha}\$$

where $$\\alpha\$$ is the angle between the plain of the coil itself and and the Lorentz force (which is slightly different than 90 degrees in the configuration of the previous picture.)

Now the equation $$\\vec{T} = \vec{m}\times\vec{B}\$$ is well satisfied but, as I previously mentioned, when the flux $$\\phi(\vec{B})\$$ is at its maximum than the torque $$\\vec{T}\$$ is zero, while when the flux $$\\phi(\vec{B})\$$ is zero than the torque $$\\vec{T}\$$ is proportional to product of $$\i\$$.

Is there anyone who can clarify me this issue, please?

• "when the flux is at its maximum then the torque is zero..." This is incorrect. How do you arrive at this notion? Feb 24, 2021 at 19:12
• In the picture above the torque T is almost at its maximum, while the flux is almost equal to zero. Feb 24, 2021 at 19:47
• The flux is not zero and is constant. It's not flux passing through the middle of the winding that produces torque, but the interaction of the flux and the current in the winding Feb 24, 2021 at 19:58
• the flux $\phi$ of a constant vector through a planar surface S, such as the magnetic field $\vec{B}$ across a rectangular closed loop, is defined as the scalar product of magnetic field $\vec{B}$ and a vector, called the surface vector $\vec{S}$, which is perpendicular to the surface and has a length equal to the area of the surface itself. That scalar product, in the picture above, is clearly 0. Consequently I can accept $T\propto\phi$ only if we refer to $\phi$ as the maximum flux passing through the coil while it is rotating around its axis. Feb 24, 2021 at 20:56
• There is no "surface" in the Lorentz equation Feb 24, 2021 at 21:30

You need to add commutator to your model to rectify torque direction. so we got

$$\T = ilbB|sin{\alpha}|\$$

then average torque will equals

$$\T_{avg} = \frac{ilbB}{2\pi} \int_{0}^{2\pi}sin(\alpha)\;d\alpha\$$

$$\T_{avg} = \frac{4ilbB}{2\pi}\$$

which mean average torque is proportional to current. You can do with 3 phase motor as well.

For magnetic flux.

$$\\phi=B\times A\$$ For A is area,from

$$\T = ilbB|sin{\alpha}|\$$ when b is distant between 2 wire and l is length $$\l\times b\$$ will equals to A so $$\T = iAB|sin{\alpha}|\$$ and $$\T = i\phi|sin{\alpha}|\$$

So in conclusion.

$$\T \propto \phi i\$$

• ok, that means average T is proportional to the current, which is half of the sentence I've mentioned, the other one being T proportional to the magnetic flux, which does not seem to be true Feb 24, 2021 at 19:45
• I add that part to my answer. Are you notice that length times width equals to area, this surprise me at first time Feb 24, 2021 at 20:08
• It looks like we are close to the end, just a further clarification... when you say $\phi = B\times A$ you use $\phi$ for the maximum magnetic flux available during the rotation. In that case you can write $T\propto\phi i$, which should provide the definitive answer to my question Feb 24, 2021 at 20:35
• Yes, seem like ϕ is just simplification for variable nothing more. Feb 24, 2021 at 21:21
• I just add conclusion as you wish. Hope thes is answer you seek. Feb 24, 2021 at 21:28

From conservation of energy, you know that electrical input power is equal to mechanical output power plus losses. Voltage times current is equal to torque times speed plus losses. Speed is proportional to voltage and torque is proportional to current. A good text will show the derivation.

Both cases (case of simple loop in 1st case and dc rotor with only single loop ) are slightly different and difference is very subtle ! And that subtlty causes a fundamental difference which you cannot figure it out -lets talk about that difference -

But before that I'll assume some basic assumption for simplicity-

1.Dc motor is of two pole only .

2.only one simple conducting loop is placed in rotor's slot

3 . Rotor is always under the Pole faces.

4 . Commutator is placed accordingly.

In first case(that you mentioned) - flux is varying sinusoidally i.e $$\phi=K Sin\omega t$$ and torque also vary sinusoidally (ignoring back EMF )i.e $$T=P Cos\omega t$$ and due to phase difference of the π/2 between Flux and torque , when one quantity become Zero other become maximum (magnitude)

In second case- under the above 4assumption that I assumed - flux is varying linearly i.e $$[\phi=Kt ] for 0<\theta<π/2$$ and after π/2 flux start decreasing linearly and so on ......

And $$torque = constant$$ because angle between 'l' and 'B' is constant and hence you wouldn't get max or min torque !