AngelTyrael said:
Hey there, I've been searching online for a quite a decent amount of time trying to wrap my head around this concept.
We've been asked to research the link between Lenz's law and DC motors, and for some reason I was given the impression that back emf was a positive to DC motors.
After some online research, all that I've concluded is that back emf is produced when the voltage is increased, which leads to the motor operating at its maximum capacity.
What is the most obvious effect of Lenz's law and the way in which DC motors function?
Thanks in advance.
Lenz's law is when you pass magnetic field lines across a closed circuit and thereby inducing a current within it. The current does not rise instantaneously with respect to the voltage, there is a delay. The property of the circuit that determines the delay with respect to the voltage is its inductance. In circuits without stray capacitance, we have the following equation:
V_{induced}=V_{resistance}+V_{inductance}
Where:
V_{induced}, is the induced voltage, equal to rate change of magnetic flux per unit time; "webers per second" in SI Units.
V_{resistance}=RI
R, is the resistance.
I, is the current.
V_{inductance}=L\frac{dI}{dt}
L, is the inductance.
\frac{dI}{dt}, is the change of current per change in time.
Thus, a magnet approaching a circuit will induce currents in the circuit that produce magnetic fields that tend to repel the magnet. A magnet receding from a circuit also produces currents in it, but in such a way that tends to attract the magnet.
A classic example:
One very strong magnet is dropped through a tube of aluminum. The magnet induces current inside the tube of aluminum that acts to slow down its movement. As the magnet falls through, the currents in the aluminum tubing above the magnet pull the magnet towards it, in an upwards direction. The currents in the tubing below pushes the magnet away from it, also in a upwards direction.
A sufficiently strong magnet can generate a visibly noticeable drag on the magnet. The sheet of aluminum, compared to other materials, produces much magnetic energy for a given induced voltage. This occurs because aluminum has a low resistivity, which means it has high conductivity.
Given:
V_{resistance}=RI
And because:
R=\rho\frac{\ell}{A}
Where:
\rho, is the resistivity.
\ell, is the length of the path of conductivity.
A, is the cross-sectional area of the path of conductivity.
Resistivity can therefore be expressed as the following:
\rho=\frac{V_{resistance}}{I}\frac{A}{\ell}
\rho=\frac{V_{induced}-V_{inductance}}{I}\frac{A}{\ell}
A low resistivity \rho in a conducting path with a specified length and cross-section results in a greater I, for a given V_{resistance}.
Magnetic energy stored in a field due to these currents is defined as E_{mag}=\frac{1}{2}LI^2. Isolating I in the equation results in the equation I=\sqrt{\frac{2E_{mag}}{L}}. The equation for \rho becomes:
\rho=\left(V_{resistance}\right)\sqrt{\frac{L}{2E_{mag}}}\frac{A}{\ell}
Therefore, the magnetic energy stored in the medium, in relation to the resistivity and dimensions of the conductive path, as well as any associated voltages equals:
E_{mag}=\frac{1}{2}\left(\frac{V_{resistance}}{\rho}\sqrt{L}\frac{A}{\ell}\right)^2
Proving that the energy in the magnetic field produced as a result of back EMF is enhanced by the presences of low resistivity materials. To see this, see what happens to E_{mag} if \rho is decreased. The equation is also helpful in proving that using a thicker cross-section of the conduction path enhances the energy in the opposing magnetic field.