How to find the length of the wire?

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The discussion focuses on calculating the length of wires needed for a radio signal with a wavelength of 2 meters at 150 MHz. It is clarified that each wire should ideally be a quarter wavelength, resulting in a length of 0.5 meters for a half-wave dipole antenna. A participant initially suggested a wire length of 0.25 meters, which was corrected to reflect the proper relationship between wavelength and wire length. The importance of including units in calculations to avoid confusion is emphasized, alongside the understanding that the overall antenna length should be half the wavelength. The conversation highlights the principles of antenna design and the relationship between frequency and wavelength.
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Homework Statement


For a 150 MHz (1 MHz=1 million cycles per second) radio signal, the wavelength is 2 m long. If each wire is about 0.5 m long, a half wavelength of radiation will span the two wires taken as one. The result is that electromagnetic waves are generated in the direction of the arrow, moving away from the transmitter at the speed of light.

Homework Equations


To transmit radiation of wavelength 1, how long should each wire be in multiples of (or fraction of) 1?

The Attempt at a Solution


I'm not sure if this is the answer, but I was thinking the length of the wire could be 0.25m if the wavelength is 1? I don't really understand though, could someone help please? Thanks.
 
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I'd agree with your answer.
How did you decide on that?
 
lulu said:

Homework Statement


For a 150 MHz (1 MHz=1 million cycles per second) radio signal, the wavelength is 2 m long. If each wire is about 0.5 m long, a half wavelength of radiation will span the two wires taken as one. The result is that electromagnetic waves are generated in the direction of the arrow, moving away from the transmitter at the speed of light.

Homework Equations


To transmit radiation of wavelength 1, how long should each wire be in multiples of (or fraction of) 1?

The Attempt at a Solution


I'm not sure if this is the answer, but I was thinking the length of the wire could be 0.25m if the wavelength is 1? I don't really understand though, could someone help please? Thanks.
Welcome to the PF.

Could you use the Upload button (in the lower right of the Reply window) to upload a PDF or JPEG of the figure that goes with this problem? I'm having trouble understanding the problem statement. Thanks. :smile:
 
berkeman said:
Welcome to the PF.

Could you use the Upload button (in the lower right of the Reply window) to upload a PDF or JPEG of the figure that goes with this problem? I'm having trouble understanding the problem statement. Thanks. :smile:

Thanks, this was the only photo that was given to refer to.
 

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Merlin3189 said:
I'd agree with your answer.
How did you decide on that?

Thanks for replying. I'm not sure if I'm thinking about this correctly but I thought of it as a proportion problem. 0.5/2= x/1; so x=0.25.
 
lulu said:
Thanks, this was the only photo that was given to refer to.
So it's a typical half-wave dipole antenna (2x Quarter Wave elements).
lulu said:

Homework Statement


For a 150 MHz (1 MHz=1 million cycles per second) radio signal, the wavelength is 2 m long. If each wire is about 0.5 m long, a half wavelength of radiation will span the two wires taken as one. The result is that electromagnetic waves are generated in the direction of the arrow, moving away from the transmitter at the speed of light.

Homework Equations


To transmit radiation of wavelength 1, how long should each wire be in multiples of (or fraction of) 1?

The Attempt at a Solution


I'm not sure if this is the answer, but I was thinking the length of the wire could be 0.25m if the wavelength is 1? I don't really understand though, could someone help please? Thanks.
So what is the issue? At 150MHz the wavelength is 2m. A quarter wavelength is 0.5m, so it takes two of those elements to make a dipole. Where did 0.25m come from? Sorry if I'm missing something...
 
berkeman said:
So it's a typical half-wave dipole antenna (2x Quarter Wave elements).

So what is the issue? At 150MHz the wavelength is 2m. A quarter wavelength is 0.5m, so it takes two of those elements to make a dipole. Where did 0.25m come from? Sorry if I'm missing something...

So I'm trying to figure out what the length of the wire is if the wavelength is 1.
I wasn't sure how to approach it so I just did this: 0.5/2= x/1; so x=0.25
 
lulu said:
So I'm trying to figure out what the length of the wire is if the wavelength is 1.
One what? Please include units. Thanks.
 
Ah, sorry. I think I misread the question! I get the question now.
Thanks for your help!
 
  • #10
Proportion is completely valid for this topic. That is what I meant when I asked how you got your answer.
The radiator is half a wavelength and the individual 'wires' are each a quarter wavelength. So when the frequency is doubled, the wavelength is halved and all other linear measuements are halved.

Generally we don't talk about the two 'wires'. This radiator is called a "half wave dipole" because its overall length is half a wavelength. The current is supplied at the centre, so there is a gap there and each side is then 1/4 λ.
For small wavelengths the wires are usually rigid metal tubes (often aluminium) such as you may see as radio or TV antennae on houses. (They also have extra tubes as well as the dipole.)

Incidentally, the diagram shows the radiation at right angles to the wires, which is correct. But that also means down, towards you, away from you and at any direction which is at right angles to the wires.at right angles. Also this is merely the direction of maximum radiation. There is also some radiation at other angles. The only direction in which there should be no radiation is directly in line with the wires. (Maybe too much info for now. Just that on PF we do try to be correct.)

As berkeman says, normally give units with your measurements - it avoids confusion, both for us and more importantly for you.
Here, because it is a question of proportion, you could reasonably give answers without units (as in post #7.) If the wavelength is 1 then the 'wires' are 0.25 and it does not matter whether we mean 1m and 0.25m or 1ft and 0.25ft or 1dm and 0.25dm or 1 perch and 0.25 perch, etc. so long as the units are the same.
 
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