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field expresion question..

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http://i29.tinypic.com/2rdiuye.jpg

i cant understand how they got the first expression

in a normal coil the distance is r
but here they have a root in the denominator
like they used Pythagorean theorem and multiplying by sinus

and i cant understand where they do it here in order to get the
field expession
??
 
in a normal coil the distance is r
but here they have a root in the denominator
like they used Pythagorean theorem and multiplying by sinus

and i cant understand where they do it here in order to get the
field expression
??
 

skyhawk

New Member
The distance from the field point (on the axis) to a point on the coil is

√(x^2 + R1^2)

Write dl and r in cylindrical coordinates, do the cross product, and integrate.
 
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i have this formula for a magnetic inside a coil

[latex]
B=\frac{\mu _0NI}{l}
[/latex]

this is a cross section of the system.
to have a mutual induction
i would need to have some current in one of the wires

i dont have it here

??
 
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skyhawk

New Member
The sample solution has a very good explanation. Read it. Understand it.

The most time consuming part is calculating the field due to the large coil using B-S law. It's not hard. Write the vectors in cylindrical coordinates, do the cross product, and integrate. I did it for myself yesterday before posting my reply. The integrals are easy. Setting up the vectors is the most difficult part.
 
the solution doesnt have a sketch.
and as i see it
the bio savart formula for inside a coil
doesnt have a distance element
[latex]
B=\frac{\mu _0NI}{l}
[/latex]

so cant see how distance can be involved in here?

the large coil cannot produce a field inside of itself
which has a distance

??
 

skyhawk

New Member
These coils are not solenoids. They are "flat" coils on the same axis spaced x apart.

Do the cross product and the integrals, and you will get the first expression in the sample answers!
 
i cant imagine them as flat 2d circles because
i was told that they have N1 and N2 turns each

if it were a 2d circles then i know how to get to its formula
from biot savart.

but here it is not because i am given that they have turns
thus making them 3d object

which quite remind me of a selonoid??

i cant understand how it both flat and has turs??
 

skyhawk

New Member
You don't know how large N1 and N2 are. Maybe they are small numbers 2 or 3 or 10 or so, and the diameter of the wire is small compared to the other dimensions; therefore this is a good approximation. If the coils were solenoids you would need the length which you weren't given. Besides in an ideal solenoid there is no field outside the solenoid, so how could one coil produce a flux in the other coil if they are x apart? The geometric description just doesn't fit for solenoids. I've told you how to get the answer using the B-S law that you've studied. Do what I've told you, and you will get the answer. If you have other questions ask your instruct.
 
ok i presume that in coil1 there is circuling electricity
which creates flux in coil 2.

[latex]B=\oint \frac{\mu_0}{4\pi}\frac{Idl\times \hat{r}}{r^2}\\[/latex]
[latex]dl=rd\theta\\[/latex]
[latex]B=\int_{0}^{2\pi} \frac{\mu_0}{4\pi}\frac{Idl}{r^2}=\int_{0}^{2\pi} \frac{\mu_0}{4\pi}\frac{Ird\theta}{r^2}=\frac{\mu_0}{4\pi}\frac{I2\pi}{r}\\[/latex]
[latex]\phi_1=B_1A_1=\frac{\mu_0}{4\pi}\frac{I2\pi}{R_1N_1}\pi (R_1)^2=\frac{\mu_0}{4}\frac{I2\pi}{N_1} R_1\\[/latex]
[latex]M_{12}=\frac{N_2\phi_{12}}{I1}\\[/latex]
[latex]M_{12}=\frac{N_2\phi_1}{I_1}=\frac{N_2\frac{\mu_0}{4}\frac{I2\pi}{N_1} R_1}{I_1}[/latex]

but its not like in the solution at all
and i tried to stick by the book
where is the mistake?
 
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skyhawk

New Member
but its not like in the solution at all
and i tried to stick by the book
where is the mistake?
You didn't do a correct cross product. The cross product of two vectors is another vector. I'm already written the vectors dl and r that work.

r is the distance from a point on the coil to the field point. It is not the same as the radius of the coil, which is R1.
 
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cross product of dl (if i am going counter clockwise)
and r (pointed to the center
so by right hand rule
i get vector pointer to the z axes (0,0,1) vector
[latex]B=\oint \frac{\mu_0}{4\pi}\frac{Idl\times \hat{r}}{r^2}\\[/latex]
[latex]dl=rd\theta\\[/latex]
[latex]B=\int_{0}^{2\pi} \frac{\mu_0}{4\pi}\frac{Idl}{r^2}=\int_{0}^{2\pi} \frac{\mu_0}{4\pi}\frac{Ird\theta}{r^2}=\frac{\mu_0}{4\pi}\frac{I2\pi}{r}\hat{z}\\[/latex]
dl and r always perpandicular
so their cros product is always 1.
i tried to visualizebmy thoughts in this drawing
http://i31.tinypic.com/radu8j.jpg
where is my mistake in how i imagine the system
 

skyhawk

New Member
The field point is not at the center of the coil. It is on the axis a distance x from the plane of the coil. In previous posts I have given you everything you need to solve the problem.
 
sorry but my english is not very good

can you tell me what
field point means
??

i am trying to solve it all the time
 
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The equation you wrote is not the Biot-Savart Law.

See here:

Biot?Savart law - Wikipedia, the free encyclopedia

dl = ds(-i sinθ + j cosθ)

r = -R1(i cosθ + j sin) + x k

r = √(x^2 + R1^2)

ds = R1dθ

Plug in and do the integral.
i noticed that for each dl it has a different [latex]\hat{r}[/latex]

how you came up with this expression

r = -R1(i cosθ + j sin) + x k

i tried to break it into components like you did
http://i26.tinypic.com/9sz51l.jpg
but i dont know where to put theta
??

i tried to imagine r where its not tangent in the middle of dl (i think its wrong)
in here dtheta called x
http://i28.tinypic.com/2ia9ank.jpg

and still i get rcos(x) i-rsin(x)j

and its still differs you expression
i took r as direction unit vector
but you took R1 radu
why??
where is my mistake
??
 
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