Would the reading in voltmeter change?

[Daniel Han]

I'm Referencing: Foundations of Electromagnetic Theory - 4th Edition by Reitz

Here's where I'm struggling:

-an electric field will exist in the wire, the field being related to $$\Delta \phi$$ by the relation

$$\Delta \phi = \int \mathbf{E} \cdot d\mathbf{l}$$ (7-12a)

...

Furthermore, because of the geometry, the electric field must be the same at all points along the wire. Therefore, Eq (7-12a) reduces to

$$\Delta \phi = El$$

Then take a look @ this crude diagram of a circuit I've drawn.

A voltmeter probe marked P (red) is moving along the circuit (red arrow), while the black probe remains stationary.

Then would the reading in Voltmeter change?

According to the quote, it should. Because assume the the two probes started @ one end of the battery's terminal. Then, the probe P, would end up @ the opposite terminal after its travel around the circuit, thus V = 0, should change into V = V_0.

Is this correct?

 

[Svein]

Then would the reading in Voltmeter change?

Possibly. But I think you would get nasty burn wounds and, depending on the capacity of the battery, the wire would melt.

 

[Redbelly98]

I'm Referencing: Foundations of Electromagnetic Theory - 4th Edition by Reitz

Here's where I'm struggling:
Then take a look @ this crude diagram of a circuit I've drawn.

A voltmeter probe marked P (red) is moving along the circuit (red arrow), while the black probe remains stationary.

Then would the reading in Voltmeter change?

According to the quote, it should. Because assume the the two probes started @ one end of the battery's terminal. Then, the probe P, would end up @ the opposite terminal after its travel around the circuit, thus V = 0, should change into V = V_0.

Is this correct?

Yes, you are correct. We are not treating the wire as ideal in this case, but allowing for the fact that it has some resistance, and does not short-circuit the battery -- otherwise, as Svein said, burns and melting wire become a concern.

So: do not try this with a normal wire connected to a battery!

 

[Daniel Han]

Yes, you are correct. We are not treating the wire as ideal in this case, but allowing for the fact that it has some resistance, and does not short-circuit the battery -- otherwise, as Svein said, burns and melting wire become a concern.

So: do not try this with a normal wire connected to a battery!

Thanks for the response! But with Svein's answer, why would I get burned/wire melt? Is the presumed phenomenon dependent on the presence of the Voltmeter?

I'm not quite getting where the thermal energy would possibly come from.

 

[sophiecentaur]

Thanks for the response! But with Svein's answer, why would I get burned/wire melt? Is the presumed phenomenon dependent on the presence of the Voltmeter?

I'm not quite getting where the thermal energy would possibly come from.

Presumably you are coming to this problem with a certain about of knowledge of the common formulae used in electrical calculations. If the wire has finite resistivity then each cm of the wire will have a resistance and you can replace your wire with a number of resistors connected in series. The current (I) through the circuit will be V/(total R) and the Voltage drop across any of the resistors will be I times that resistance. IS there much more to say about the situation, once it has been stated that way?

If you want to go deeper into the 'meaning' of resistance then it is merely the ratio of Voltage across and Current through a component. If you use the Maths to work out things from the basic definitions then it all comes out very easily. If you want an explanation of how energy is transferred from a Charge to the lattice of a metal it is passing through, then the Physics can be addressed on many levels but energy is needed to cause the charges to be moved and that can turn up as internal energy (heat) or as Mechanical work, in the case of an electric motor.

 

[Daniel Han]

Presumably you are coming to this problem with a certain about of knowledge of the common formulae used in electrical calculations. If the wire has finite resistivity then each cm of the wire will have a resistance and you can replace your wire with a number of resistors connected in series. The current (I) through the circuit will be V/(total R) and the Voltage drop across any of the resistors will be I times that resistance. IS there much more to say about the situation, once it has been stated that way?

If you want to go deeper into the 'meaning' of resistance then it is merely the ratio of Voltage across and Current through a component. If you use the Maths to work out things from the basic definitions then it all comes out very easily. If you want an explanation of how energy is transferred from a Charge to the lattice of a metal it is passing through, then the Physics can be addressed on many levels but energy is needed to cause the charges to be moved and that can turn up as internal energy (heat) or as Mechanical work, in the case of an electric motor.

Oh, the resistance, then yes. But I have played around with circuits with battery connected to wires and etc, etc... but never have encountered a problem when a "normal wire" (as stated by Redbelly98) melted, or got a severe burn. And I understood the responses that the wire-melting/burning is, in fact, a common occurrence.

Because if it's a normal conducting wire with low resistance, merely by connecting to a battery shouldn't be a problem.

It feels like I'm missing something.

 

[davenn]

Because if it's a normal conducting wire with low resistance, merely by connecting to a battery shouldn't be a problem.

It feels like I'm missing something.

because if it is a normal wire connected between the ends of the battery a lot of current will flow

say the battery is 10V and the wire is 1 metre ( 3ft) long, say the resistance of the wire of that length is 1 Ohm

current flowing = I = V / R
10V / 1 Ohms = 10 Amps of current flow ( assuming the battery can supply 10A ( it will supply as much as it is able to))

power dissipated by the wire in Watts =
W = V x I = 10V x 10 A = 100Wthe wire is going to get VERY hot and will probably burn out like a fuse wireDave

 

[Daniel Han]

because if it is a normal wire connected between the ends of the battery a lot of current will flow

say the battery is 10V and the wire is 1 metre ( 3ft) long, say the resistance of the wire of that length is 1 Ohm

current flowing = I = V / R
10V / 0.1 Ohms = 10 Amps of current flow ( assuming the battery can supply 10A ( it will supply as much as it is able to))

power dissipated by the wire in Watts =
W = V x I = 10V x 10 A = 100Wthe wire is going to get VERY hot and will probably burn out like a fuse wireDave

Now that was an answer I was looking for! Thanks!

 

[TurtleMeister]

Thanks for the response! But with Svein's answer, why would I get burned/wire melt? Is the presumed phenomenon dependent on the presence of the Voltmeter?

I'm not quite getting where the thermal energy would possibly come from.

I agree with Dave's post #7, but I would just like to add that such an experiment is potentially dangerous. Batteries have an internal resistance and when short circuited most of the energy will be released by the battery itself (if the wire size can handle the current). Batteries with a high energy density, such as lithium polymer, could burn or explode releasing toxic fumes.

Lower energy density batteries or not fully charged batteries may only get hot, but it's still not a good idea to short circuit them.

 

[sophiecentaur]

but never have encountered a problem when a "normal wire" (as stated by Redbelly98) melted, or got a severe burn.

This will be because you have never used a large enough battery which will give enough current. All batteries have internal resistance, which limits the amount of current available for a given output voltage. Your experiments may well have resulted in partial damage or destruction of your batteries. Apply a bit of theory before trying random experiments and you will save money (and fingers).