# Magnetic flux through a triangular loop?

• Abdulwahab Hajar
In summary, the mutual inductance between a very long, straight wire and a conducting equilateral triangular loop is B = µI/2πr. The Attempt at a Solution found that the magnetic flux through the triangle is Φ = ∫B.dA. The bounds for the height of the triangle at different distances from the wire can be found by solving a linear equation in one variable, r.
Abdulwahab Hajar

## Homework Statement

Determine the mutual inductance between a very long, straight wire and a conducting equilateral triangular loop

## Homework Equations

I know how to do it but I'm stuck at the step of finding the total inductance through the triange.

## The Attempt at a Solution

we know that the magnetic field due to an infinitely long wire is
B = µI/2πr
in order to find the magnetic flux through the triangle we need to find Φ = ∫B.dA
we have B but what would dA be in this case
the book says it's (2/√3) * (r - d) *dr
why is that so?

thank you

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Look at the geometry for your triangle.
Starting at the point closest to the wire, we will call that ##(d, 0)## , you will follow a counterclockwise path to ##(d+\frac{b\sqrt{3}}{2}, -\frac b2 )##, then from
##(d+\frac{b\sqrt{3}}{2}, -\frac b2 )## to ##(d+\frac{b\sqrt{3}}{2}, +\frac b2 )##, and finally from ##(d+\frac{b\sqrt{3}}{2}, +\frac b2 )## to ##(d,0 )##.
Of course, you can parameterize each of those line integrals to get a form that is easier to work with.

Abdulwahab Hajar
RUber said:
Look at the geometry for your triangle.
Starting at the point closest to the wire, we will call that ##(d, 0)## , you will follow a counterclockwise path to ##(d+\frac{b\sqrt{3}}{2}, -\frac b2 )##, then from
##(d+\frac{b\sqrt{3}}{2}, -\frac b2 )## to ##(d+\frac{b\sqrt{3}}{2}, +\frac b2 )##, and finally from ##(d+\frac{b\sqrt{3}}{2}, +\frac b2 )## to ##(d,0 )##.
Of course, you can parameterize each of those line integrals to get a form that is easier to work with.
Ok that really helped to be honest.
Thank you sir
But that would be a line integral right?
what if I wanted to cover the surface as in the area covered by the sides of the triangle

Then, in that case, you would have x going from d to d+b/2 * sqrt(3) and y ranging from the lower line to the upper line (as a function of x).
These bounds are no fun to work with, but will get you to the answer for the surface integral.

Abdulwahab Hajar
RUber said:
Then, in that case, you would have x going from d to d+b/2 * sqrt(3) and y ranging from the lower line to the upper line (as a function of x).
These bounds are no fun to work with, but will get you to the answer for the surface integral.
Yeah I get ya...
is there anyway to do it with a change in r since it covers both x and y directions...
this is how it is in the solution manual perhaps if you take a look you could understand how it was derived
I just couldn't figure it out

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I see what they did there. Since the wire in infinitely long, B is only dependent on r, the distance from the line.
So, if you slice up the equilateral triangle vertically, along lines with the same distance to the wire, you get
##\int\int B \, dA = \int B h(r) dr, ##
where ##h(r)## is the height of the triangle at distance r.
At r=d, the height is 0, and at r = ##d+b\sqrt{3}/2##, the height is b. So you can make a linear rule for the height that looks like
## h = (r-d)\frac{2}{\sqrt{3}} .##
Check that against the reference points to make sure.
So, now you can treat the integral in terms of one variable, r.
##\int B h(r) dr = \int B (r-d)\frac{2}{\sqrt{3}}\, dr ##
The final answer applies a simplification in the natural log, but otherwise, you should be able to follow the rest from there with about 2 extra steps.
I hope this helps.

Abdulwahab Hajar
RUber said:
I see what they did there. Since the wire in infinitely long, B is only dependent on r, the distance from the line.
So, if you slice up the equilateral triangle vertically, along lines with the same distance to the wire, you get
##\int\int B \, dA = \int B h(r) dr, ##
where ##h(r)## is the height of the triangle at distance r.
At r=d, the height is 0, and at r = ##d+b\sqrt{3}/2##, the height is b. So you can make a linear rule for the height that looks like
## h = (r-d)\frac{2}{\sqrt{3}} .##
Check that against the reference points to make sure.
So, now you can treat the integral in terms of one variable, r.
##\int B h(r) dr = \int B (r-d)\frac{2}{\sqrt{3}}\, dr ##
The final answer applies a simplification in the natural log, but otherwise, you should be able to follow the rest from there with about 2 extra steps.
I hope this helps.
It's great man
Thanks a lot
You're the best

## 1. What is magnetic flux through a triangular loop?

Magnetic flux through a triangular loop is a measure of the total magnetic field passing through the loop. It is calculated by taking the dot product of the magnetic field and the area vector of the loop.

## 2. How is magnetic flux through a triangular loop different from other shapes?

The magnetic flux through a triangular loop is different from other shapes because the area vector of a triangle is not constant, unlike a rectangle or circle. This means that the angle between the area vector and the magnetic field is constantly changing, resulting in a varying magnetic flux.

## 3. How is the direction of the magnetic flux through a triangular loop determined?

The direction of the magnetic flux through a triangular loop is determined by the right-hand rule. If the fingers of the right hand are curled in the direction of the magnetic field, the thumb will point in the direction of the magnetic flux.

## 4. What factors affect the magnetic flux through a triangular loop?

The magnetic flux through a triangular loop is affected by the strength and direction of the magnetic field, as well as the orientation of the loop in relation to the magnetic field. It is also affected by the size and shape of the loop.

## 5. How is the magnetic flux through a triangular loop used in practical applications?

The magnetic flux through a triangular loop is used in various practical applications, such as in generators and electric motors. It is also used in magnetic sensors and detectors, such as in compasses and metal detectors.

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