Magnetic field at the center of 3 wires carrying current?

[Charles Link]

By symmetry the field due to A can be said to be canceled by one of the other fields B or C. That means you only have to find the field due to either C on its own or B on its own. Both fields have the same magnitude and direction at the point in question.

It's just a shortcut suitable for this particular symmetrical set up only. I would advise you to proceed with the other methods being outlined here because they can deal with all arrangements.

At point $O$, the magnetic field from $A$ does not cancel the magnetic field from $B$, nor does it cancel the magnetic field from $C$. $\\$ It may seem logical to do this, but it is incorrect to think of the magnetic field at $O$ from anyone of $A$, $B$, or $C$ as being clockwise or anticlockwise. The magnetic fields around a given wire of current are clockwise or anti-clockwise, but when looking at a point $O$, the magnetic field from anyone of $A$, $B$, or $C$ is a vector that points in a given direction in the x-y plane, rather than clockwise or anti-clockwise. $\\$ There is a shortcut solution that I presented in the first line or two of post 16, which I will describe in more detail here: If the currents are all in the same direction (e.g. out of the paper) from the three wires, by symmetry the magnetic fields from them will cancel at $O$. This is because if there were a non-zero component at $O$ from this symmetric configuration, it would need to also point in the two other directions (120 degrees apart) by symmetry. Thereby, three wires with currents in the same direction gives the result of zero for the magnetic field at $O$. (And then the current from wire $A$, which is one unit of current into the paper, can be considered to be composed of one unit of current out of the paper, along with two units into the paper). $\\$ The magnetic field at $O$ from the two units of current into the paper at $A$ is then the only thing that needs to be computed to get the answer to the total magnetic field at $O$ from the currents in the three wires for the problem as it is given.

 

[Dadface]

At point $O$, the magnetic field from $A$ does not cancel the magnetic field from $B$, nor does it cancel the magnetic field from $C$. $\\$ It may seem logical to do this, but it is incorrect to think of the magnetic field at $O$ from anyone of $A$, $B$, or $C$ as being clockwise or anticlockwise. The magnetic fields around a given wire of current are clockwise or anti-clockwise, but when looking at a point $O$, the magnetic field from anyone of $A$, $B$, or $C$ is a vector that points in a given direction in the x-y plane, rather than clockwise or anti-clockwise. $\\$ There is a shortcut solution that I presented in the first line or two of post 16, which I will describe in more detail here: If the currents are all in the same direction (e.g. out of the paper) from the three wires, by symmetry the magnetic fields from them will cancel at $O$. This is because if there were a non-zero component at $O$ from this symmetric configuration, it would need to also point in the two other directions (120 degrees apart) by symmetry. Thereby, three wires with currents in the same direction gives the result of zero for the magnetic field at $O$. (And then the current from wire $A$, which is one unit of current into the paper, can be considered to be composed of one unit of current out of the paper, along with two units into the paper). $\\$ The magnetic field at $O$ from the two units of current into the paper at $A$ is then the only thing that needs to be computed to get the answer to the total magnetic field at $O$ from the currents in the three wires for the problem as it is given.

You're right that I was wrong Charles Link. So thank you. You will get cancellation between two wire carrying equal currents but only at the mid point of those wires and not at point O as the question. I didn't spend enough time looking at the question and jumped in too quickly. If that wasn't bad enough I also made a mistake in using the corkscrew rule and the cancellation I referred to occurs for currents flowing in the same direction. Bad mistakes and I'm blaming that on the curry I had last night. Sorry Helly 123.

One way I'm thinking of the question now is to draw a plan view of the arrangement shown in the question.

1. Draw a line from A to O.

2. Draw a second line which passes through O and which is at 90 degrees to the first line. This second line will be tangential to the circular field line at O which would be due to A on its own.

2.Use whatever rule you use to find the direction of the field at O. I use the corkscrew rule...sometimes incorrectly it seems.

3. Repeat for the other two wires.

4. Calculate B for one of the wires. B will have the same value (but different directions) for all three wires.

5.Use a bit of geometry and a bit of vector addition to calculate the resultant B.

 

[Helly123]

At point $O$, the magnetic field from $A$ does not cancel the magnetic field from $B$, nor does it cancel the magnetic field from $C$. $\\$ It may seem logical to do this, but it is incorrect to think of the magnetic field at $O$ from anyone of $A$, $B$, or $C$ as being clockwise or anticlockwise. The magnetic fields around a given wire of current are clockwise or anti-clockwise, but when looking at a point $O$, the magnetic field from anyone of $A$, $B$, or $C$ is a vector that points in a given direction in the x-y plane, rather than clockwise or anti-clockwise. $\\$ There is a shortcut solution that I presented in the first line or two of post 16, which I will describe in more detail here: If the currents are all in the same direction (e.g. out of the paper) from the three wires, by symmetry the magnetic fields from them will cancel at $O$. This is because if there were a non-zero component at $O$ from this symmetric configuration, it would need to also point in the two other directions (120 degrees apart) by symmetry. Thereby, three wires with currents in the same direction gives the result of zero for the magnetic field at $O$. (And then the current from wire $A$, which is one unit of current into the paper, can be considered to be composed of one unit of current out of the paper, along with two units into the paper). $\\$ The magnetic field at $O$ from the two units of current into the paper at $A$ is then the only thing that needs to be computed to get the answer to the total magnetic field at $O$ from the currents in the three wires for the problem as it is given.

At point $O$, the magnetic field from $A$ does not cancel the magnetic field from $B$, nor does it cancel the magnetic field from $C$. $\\$ It may seem logical to do this, but it is incorrect to think of the magnetic field at $O$ from anyone of $A$, $B$, or $C$ as being clockwise or anticlockwise. The magnetic fields around a given wire of current are clockwise or anti-clockwise, but when looking at a point $O$, the magnetic field from anyone of $A$, $B$, or $C$ is a vector that points in a given direction in the x-y plane, rather than clockwise or anti-clockwise. $\\$ There is a shortcut solution that I presented in the first line or two of post 16, which I will describe in more detail here: If the currents are all in the same direction (e.g. out of the paper) from the three wires, by symmetry the magnetic fields from them will cancel at $O$. This is because if there were a non-zero component at $O$ from this symmetric configuration, it would need to also point in the two other directions (120 degrees apart) by symmetry. Thereby, three wires with currents in the same direction gives the result of zero for the magnetic field at $O$. (And then the current from wire $A$, which is one unit of current into the paper, can be considered to be composed of one unit of current out of the paper, along with two units into the paper). $\\$ The magnetic field at $O$ from the two units of current into the paper at $A$ is then the only thing that needs to be computed to get the answer to the total magnetic field at $O$ from the currents in the three wires for the problem as it is given.

Thanks Charles Link. Btw, the answer on my textbook is weird. 18.4 T which is illogical for me.

 

[Charles Link]

Thanks Charles Link. Btw, the answer on my textbook is weird. 18.4 T which is illogical for me.

There is now a general agreement on this one for what the correct answer is. We aren't supposed to tell you the answer, but please tell us what you computed, and I will let you know if @vanhees71 , @QuantumQuest , and myself all agree with that answer.

 

[Helly123]

There is now a general agreement on this one for what the correct answer is. We aren't supposed to tell you the answer, but please tell us what you computed, and I will let you know if @vanhees71 , @QuantumQuest , and myself all agree with that answer.

Yes. My calculation
Total B = 2B = 2 $\frac{u*I}{2 \pi r}$ = $2.3 \ 10^{-5} T$

 

[Charles Link]

That is correct. Very good. I think I also see where your book (answer of 18.4 T from your post 33) went wrong. Take their answer and multiply it by $\mu_o=4 \pi \cdot 10^{-7}$ and I think you get the correct answer. Once again, very good ! :) :) $\\$ Edit: And don't forget to specify the direction which is "to the left".

 

[vanhees71]

It's amusing that nobody uses vector algebra to solve this problem once and for all but instead long winded texts. This is like in the time back when Newton in his Principia didn't reveal the public how he used his own invention, now called calculus, to present his results but pretty cumbersome geometrical constructions, which are of course ingenious but nearly incomprehensible to us today without a very detailed study of the geometrical methodology used. The reason for the use of modern vector algebra and vector calculus today (btw. thanks to Heaviside and Gibbs who developed the modern methods independently at the end of the 19th century). So I just repeat, what I wrote in the private conservation, also here in the public forum:I don't understand physics in written words well. I've no clue what is claimed by both of you. So let's calculate the magnetic field (in SI units ;-))

Each current $I$ going in positive $z$ direction produces a field
$$\vec{B}(\vec{r})=\frac{\mu_0 I}{2 \pi [(x-x_0)^2+(y-y_0)^2]}\begin{pmatrix} -(y-y_0) \\ x-x_0,0 \end{pmatrix}.$$
I've assumed an infinitely long wire here.

Working out the three vectors for the points wire and adding the three vectors gives me
$$\vec{B}(\vec{r}=0)=(-2.31 \cdot 10^{-5},0,0) \text{T}.$$

For the details see the pdf of my Mathematica notebook, I've used for the calculation

 

[Dadface]

I for one am not familiar with the method you used. Is it quicker than the method of finding the value of B at O for each wire separately and then adding the three values vectorially?

 

[vanhees71]

That's the way I did it! It basically uses Biot-Savart for each single wire and then I add up the contributions by the three wires. What's unclear with the calculation (maybe it's the uncommented Mathematica notebook)?

 

[Dadface]

I get the impression that your method is more plug and chug and probably more efficient. My preference here is to go a bit more back to first principles and sketch things out. I found that when I did a quick sketch of the vectors the answer was easy to see. However Biot and Savart and plug and chug was still necessary. I also find that it's helpful to get a more detailed visual representation of the problem. To me it becomes more obvious and you can see extra details, for example you can see straight away that if the three currents flowed in the same direction, point O would be a neutral point.

 

[vanhees71]

Then I'd draw circles around the wires (in the figure in the first posting of this thread) and then put the corresponding tangent vectors, pointing in the direction of the magnetic fields from each of the wires. In this case the magnitudes of all three fields are the same. So all vectors are of the same length. The total field is of course the vector sum, which you can also geometrically construct from this drawing. It's a lot more of work than the algebraic method, but it also helps to visualize the field.

 

[Helly123]

That is correct. Very good. I think I also see where your book (answer of 18.4 T from your post 33) went wrong. Take their answer and multiply it by $\mu_o=4 \pi \cdot 10^{-7}$ and I think you get the correct answer. Once again, very good ! :) :) $\\$ Edit: And don't forget to specify the direction which is "to the left".

Thank you. 18.4 T times 4$\pi . 10^{-7}$ = 73.6 T to the left

 

[vanhees71]

The answer is $2.31 \cdot 10^{-5} \text{T}$ to the left not 73.6 T. See my solution above:

https://www.physicsforums.com/threa...s-carrying-current.943071/page-2#post-5970872

 

[Helly123]

The answer is $2.31 \cdot 10^{-5} \text{T}$ to the left not 73.6 T. See my solution above:

https://www.physicsforums.com/threa...s-carrying-current.943071/page-2#post-5970872

yes

 

[Helly123]

This one does, I think, have a simple solution: If you write the current at A as one unit of current out of, and two units of current into the paper, the magnetic field vectors from the one unit of current out of the paper from all 3 vertices will cancel in the center, with the result that you only need to consider the problem as two units of current into the paper at A. $\\$ The alternative is to carefully sum the components of all 3 vectors. $\\$ Editing: But looking closer at it, you already solved it in the OP, but incorrectly applied the law of cosines. If you look at your vector diagram $B_{B+C}=B_{magnetic \, field}$, without the factor of $\sqrt{3}$, so that adding $B_A$ gives you $B_{total}=2 B_{magnetic \, field}$ to the left , which is what the simple method I mentioned in the first sentence gives you. You essentially solved this in the OP. $\\$(You also made an error though in your calculation of $r$ in the OP, as @QuantumQuest pointed out).

Since B for bc is B . It means the cosinus law is B^2 + C^2 - 2BC cos60 ?
Resultating in B.
So B total is B_bc + B_a = 2B to the left

 

[Charles Link]

Since B for bc is B . It means the cosinus law is B^2 + C^2 - 2BC cos60 ?
Resultating in B.
So B total is B_bc + B_a = 2B to the left

That is correct. :)

 

[Helly123]

That is correct. :)

Ok thanks