Magnetic field due to two looped wires


by physicsjock
Tags: field, looped, magnetic, wires
physicsjock
physicsjock is offline
#1
Nov25-11, 09:47 PM
P: 89
I'm just studying past exams for my physics exam in a few days.

I've been struggling with this question



The first thing i did was use the Biot-Savart law to find the magnetic field at z=0, or at ąd/2 away from z=0

I found the field for each loop seperately and since the bottom loop was orientated in the opposite direction to the above loop it was the negative of the above field. Since the dl' vector faced the opposite way when observing from z=0,

So then when i added the two fields together ofcourse they cancelled but theres no way this is right,

It says to find the derivative of B wrt z AT z=0, my equation doesnt have z in it as i found the field a distance z = ąd/2, so I suppose I could change the d/2 to z, but ill at z=0 they will still cancel get B=0 at that point so the derivative would also be zero.

In the last part it asks for the second derivative of z, with a new d, so this makes it clear my equation is wrong, haha, because mine either has d OR z

The equation i found using biot-savart for the top loop was,

B(d/2) = [itex]\frac{μIR^{2}}{2(R^{2}+\frac{d^{2}}{4})^{3/2}}[/itex]

and the bottom was just negative it


Should I be trying to find the field at an arbitrary z? If i do that, and z>d/2 then will the field be the field in the z direction of the bottom loop plus the field due to the top loop? and in the equation for the bottom loop the distance vector will have an extra d term right?

Is there an easier method I'm not seeing?

Thanks in advanced
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omoplata
omoplata is offline
#2
Nov25-11, 10:19 PM
P: 315
Yes, you HAVE to find the field at an arbitrary z since the derivatives of B(z) w.r.t. z are needed.

The magnetic field at any location is the superposition (addition) of magnetic fields due to each coil.
physicsjock
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#3
Nov25-11, 10:29 PM
P: 89
I just made the equations, I set z=0 to be z= -d/2 (in my equations)

then I had the previous equation i posted + the same equation replaced with z = (z+d) for the second loop

When i worked through that i got the first derivative was zero.

But the second derivative ive been messing with for like 40 minutes and it's no where near zero
It doesnt work because the z's are either gone, due to the second derivative, or squared due to the second derivative, product rule.

Ill attempt to write what i did haha,


B(z) = [itex]\frac{μIR^{2}}{2(R^{2}+z^{2})^{3/2}}[/itex] + [itex]\frac{μIR^{2}}{2(R^{2}+(z+d)^{2})^{3/2}}[/itex]

So the axis is set on one of the loops, so in the question where z=0 will be z=-d/2 in the equations

the first derivatives will give a 2zC + (2z+2d)C where C is all the other stuff

-dC + (-d +2d) = 0

But then the second derivative

the first two terms with the first derivative of the large part will give

2C + 2C = 4C

Then the other two terms when the second derivative is taken of the bigger thing will give

4z^2 A + (2d+2z)^2A = 2d^2 where A is the other stuff of the second derivative

and yea not zero, I expanded everything to see if there was a mistake with just leting all the other stuff = constants but I got the same result.

was what i did wrong to find the field?

I tried using amperes law, but I don't think thats the right way to do it at all, the field would be Iu/2Pir for a loop of radius r,
taking r to be z then the second loop would give B Iu/2Pi(z+d) and that doesnt seem right

omoplata
omoplata is offline
#4
Nov25-11, 11:17 PM
P: 315

Magnetic field due to two looped wires


I found the equation for B(z) on my own, and what I got is the same as what you got.

However, I don't understand how you got this,
Quote Quote by physicsjock View Post
the first derivatives will give a 2zC + (2z+2d)C where C is all the other stuff
What I got for the first derivative wasn't that simple.

I got,
[tex]\frac{\partial B(z)}{\partial z} = -\frac{3}{2} \mu I R^2 ((d+z)(R^2 + (d+z)^2)^{\frac{-5}{2}} + z(R^2 + z^2)^{\frac{-5}{2}})[/tex]

And this becomes zero when [itex]z = \frac{-d}{2}[/itex]
physicsjock
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#5
Nov26-11, 12:00 AM
P: 89
Oh cool,

was you way to find the field easier then biot-savart?

The C I set to be everything besides the (d+z) and z, because when you sub in z= -d/2 everything they are both multiplied by the same things, so i just set them to C to save worrying about it.

But yea the second derivative is giving me problems

I double checked with wolfram,
http://www.wolframalpha.com/input/?i...28-3%2F2%29%29

and its what i got, when i sub it doesnt give 0,

I get

http://www.wolframalpha.com/input/?i...shPrefs_*Math-

Thats just treating the inital terms as constant, so just a constant.


Does that mean the B is bad?
omoplata
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#6
Nov26-11, 01:41 PM
P: 315
Quote Quote by physicsjock View Post
was you way to find the field easier then biot-savart?
I used Biot-Savart.

Quote Quote by physicsjock View Post
The C I set to be everything besides the (d+z) and z, because when you sub in z= -d/2 everything they are both multiplied by the same things, so i just set them to C to save worrying about it.
Oh, OK. But keep in mind that the C includes z terms, so you can't treat C as a constant when taking the second derivative.

Quote Quote by physicsjock View Post
But yea the second derivative is giving me problems

I double checked with wolfram,
http://www.wolframalpha.com/input/?i...28-3%2F2%29%29

and its what i got, when i sub it doesnt give 0,

I get

http://www.wolframalpha.com/input/?i...shPrefs_*Math-

Thats just treating the inital terms as constant, so just a constant.


Does that mean the B is bad?
I tried this, for the case when z = 0,
http://www.wolframalpha.com/input/?i...}&a=i_Variable

And this, for the case when z = -d/2

http://www.wolframalpha.com/input/?i...}&a=i_Variable

I don't get zero either.

I'm not very familiar with Wolfram Alpha, maybe I'm doing something wrong with it.
Or maybe, like you said, there's a typo in the problem.
omoplata
omoplata is offline
#7
Nov26-11, 01:46 PM
P: 315
My links didn't display properly in the previous post. Here they are again,

z=0 case

z=-d/2 case

Edit: I figured out how to correct the links in the previous post. They work now.
physicsjock
physicsjock is offline
#8
Nov26-11, 05:24 PM
P: 89
Thanks for your help,

Yea for the second derivative i used two constants C and A.

Its weird cause like, the writer would of done the question and then found that at that weird d it =0 at the second derivative.

It doesnt make sense aswell, how could the first derivative be zero at that point and not the second,

Thanks again for your help
omoplata
omoplata is offline
#9
Nov26-11, 06:31 PM
P: 315
Quote Quote by physicsjock View Post
It doesnt make sense aswell, how could the first derivative be zero at that point and not the second,
Well, it can happen. For example, think of an object thrown upwards against gravity. It's going to fall back down. If the displacement upwards is taken to be positive, then the acceleration is -g, and remains constant for the whole duration of its motion. The velocity is initially positive as it goes up, but becomes zero when it reaches the maximum height, and then becomes negative as it falls down. At the maximum height, the velocity, which is the first derivative of displacement w.r.t. time, is zero, but the acceleration, which is the second derivative of displacement w.r.t. time, is non-zero (-g ).


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