Can I Calculate the Current and Tension Produced by a Dynamo?

In summary: I don't understand how it relates to the spinning of the rotor.The electric field is due to a time-varying magnetic field, and so the electric field at any point around a closed wire loop will be the line integral of the magnetic field along the closed loop. This line integral is always negative, because the magnetic field lines always point away from the loop. This means that the electric field will always oppose the flow of current through the wire.In order to produce a current, you need to have a load resistance (which is usually a resistor in an electrical circuit) in the circuit, and the current will flow through the load resistance and flow out the other end of the circuit. The current is proportional
  • #1
Andrea Vironda
69
3
TL;DR Summary
Continuous Current produced by a Dynamo
Hi,
I was trying to read about the current produced by a dynamo in my physics books but I can't find any reference.
I have a rotor rotating at a certain speed, air flow and with a certain inertia, but how can i find the power produced and related current and tension?
 
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  • #2
The voltage, V, produced by a dynamo is proportional to the rate of rotation = angular velocity.
The current that flows is determined by the load resistance. I = V / R .
The power; W = V * I .
The dynamo will slow down if it is overloaded.
https://en.wikipedia.org/wiki/Dynamo
 
  • #3
What you have in this case is not a voltage (potential difference) but an electromotive force. The electric field has no potential in this case since the electric field is due to a time-varying magnetic field. The electromotive force is the line integral of the electric field along the closed (!) wire loop. Let ##A## be an arbitrary surface with the wire as its boundary. Then according to Faraday's Law the electromotive force is given by the time derivative of the magnetic flux through the surface,
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
The current through the wire is then indeed given by
$$I=\frac{\mathcal{E}}{R}.$$
As you see, it is indeed crucial that we do not have a potential for the electric field here, because then the line integral along any closed path would vanish. The generator is one of the most simple example for making use of the fact that time-varying magnetic fields are always accompanied by vortices of the electric field, which in this case thus can NOT have a scalar potential.
 
  • #4
  • Can you explain better the potential topic? And vortices...

vanhees71 said:
$$\mathcal{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
  • Maybe an exponential ##^{2}## is missing in the denominator?
  • is ##\vec{f}## a versor?
 
  • #5
Do you know vector calculus? Since the thread is labeled I, I expected this to be the case.

On the right-hand side of the quoted expression there is a surface integral over the surface ##A##; ##\mathrm{d}^2 \vec{f}## are the surface-normal-vector elements. They are vectors with the length given by the area of a little surface element and a direction perpendicular to the surface. The integral is the magnetic flux through the surface, ##A##, and you take the time derivative of this magnetic flux. As defined earlier in my posting the electromotive force is defined as a line integral along the closed boundary, ##\partial A##, of the surface ##A##. The direction is chosen relative to the direction of the surface-normal elements according to the right-hand rule, i.e., pointing with the thumb of your right hand in direction of the surface-normal elements your fingers give the orientation of the boundary curve:
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}.$$
Since this line integral along a closed path is not zero in this case, there can be no scalar potential for ##\vec{E}##, because then this integral must be zero.
 
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  • #6
Yes, it's like climbing a mountain and then descending to the same point, but with a different potential.

How can I relate the spinning rotation of the disc with the produced fem? Since the produced current is proportional to the resistance, I think that I have to find a maximum of produced current as a balance between the air flow and the dynamo's resistance, otherwise I could block the rotor or let him spinning too fast without any produced current.
I brushed up the section about Faraday's law but I'm only able to find the electric field produced by a changing magnetic field on a closed loop, at a certain radius
 

Related to Can I Calculate the Current and Tension Produced by a Dynamo?

1. What is a Dynamo?

A Dynamo is a device that converts mechanical energy into electrical energy. It typically consists of a rotating coil of wire within a magnetic field, which generates an electric current.

2. How is current produced by a Dynamo?

Current is produced by a Dynamo through the process of electromagnetic induction. As the coil of wire rotates within the magnetic field, it creates a changing magnetic flux, which induces an electric current in the wire.

3. What factors affect the amount of current produced by a Dynamo?

The amount of current produced by a Dynamo is affected by several factors, including the speed of rotation, the strength of the magnetic field, and the number of turns in the coil of wire.

4. What is the difference between AC and DC current produced by a Dynamo?

AC (alternating current) is produced when the coil of wire in a Dynamo rotates continuously, while DC (direct current) is produced when the coil rotates back and forth within the magnetic field. AC is used in most power systems, while DC is used in smaller devices such as batteries.

5. Can a Dynamo produce infinite current?

No, a Dynamo cannot produce infinite current. The amount of current produced is limited by the physical properties of the materials used and the design of the Dynamo. However, advancements in technology have allowed for Dynamos to produce higher amounts of current than ever before.

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