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

Question

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?

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Answer 1

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\$

Answer 2

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.

Answer 3

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 !