# Helmholtz Coils equation

1. May 13, 2007

### stryker123

1. The problem statement, all variables and given/known data
How do I take the first and second derivatives of the helmholtz equation? The equation is as follows:

Bx= ((uR^2NI)/2) * ( [1/((x+a)^2+R^2)^(3/2) ] + [1/((x+a)^2+R^2)^(3/2) ] )

heres what I have to prove
First Derivative
http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=First.jpg [Broken]

Second Derivative
http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=Second.jpg [Broken]

2. Relevant equations
Bx= ((uR^2NI)/2) * ( [1/((x+a)^2+R^2)^(3/2) ] + [1/((x+a)^2+R^2)^(3/2) ] )

http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=First.jpg [Broken]

http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=Second.jpg [Broken]

3. The attempt at a solution
I just started multivariable calculus so I'm not quite sure what to do. I tried to take the derivative with respect to x and treated the rest of the variables as constants. However, I believe that you would need to do the chain rule to figure this problem out. Any suggestions?

Last edited by a moderator: May 2, 2017
2. May 14, 2007

### malawi_glenn

This is a one variable calculus problem. If you have taken those courses in one variable calculus, these two relations should be very familiar:

https://www.physicsforums.com/attachment.php?attachmentid=10053&stc=1&d=1179119518

I use always the first one, so this problem could be written as:

constant * [((x+a)^2+R^2)^(-3/2) + ((x-a)^2+R^2)^(-3/2)]

(you forgot one minus-sing in the denominator..)

And now you just do the thing as usual:

((x+a)^2+R^2)^(-3/2) becomes : (-3/2)(2(x+a))((x+a)^2+R^2)^(-5/2) and then the same thing for the other term. Remember to do the inner derivate (Chain-rule).

And yes, the second is just treat all other things as constants..

good luck =)

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3. May 14, 2007

### variation

Hello,

Just take first order and second order partial derivative of the equation.
I obtained the results by some calculation.
The first order derivative vanishes at x=0, which is consistent with the fact that some simple experiments reach approximate uniform magnetic field at the center of the two coils.
The second derivative does not vanish at x=0.

Regards

4. May 14, 2007

### stryker123

if i'm trying to get the equation to look like http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=First.jpg [Broken] then how can I treat the uR^2... as a constant? If i treat it as a constant, wouldn't that turn into zero?

Last edited by a moderator: May 2, 2017
5. May 14, 2007

### malawi_glenn

why?

differentiate this one for example: whith respect to x, all other is constants.

(d/dx){[(x+a)^2+ T]^(3/2)} = (3/2)*2(x+a)*[(x+a)^2+ T]^(1/2)

You should probably practice a little bit more on how to differentiate a bit more complex functions with one variable, the product rule, the chain rule etc.

Last edited: May 14, 2007
6. May 14, 2007

### malawi_glenn

did you get it now? =)

7. May 14, 2007

### stryker123

yes, I understand most of it now. Although, I am having trouble with finding the second derivative. I'm not sure if I should use the product rule or quotient rule, but I'm sure I can figure it out

Thank you to everyone who responded to this thread.

8. May 14, 2007

### malawi_glenn

you do the same thing, but now you have things like:

(d/dx){(x+a)*[(x+a)^2+ T]^(-5/2)}

And here we use the product rule OR the quotient rule, as I said earlier that I always uses the first one, it is the easiest to remember, so i always uses that one and think of the equatation(function) as a product, as x/t^3 = x*t^-3

9. May 16, 2007

### gregb

i have the same problem, and I just wanted to tell you guys this help was amazing. Thanks!

10. May 16, 2007

### malawi_glenn

=)
i hope stryker123 made it now.

it is of high imoportance to master calculus and linear algebra in the world of phyics.

11. May 16, 2007

### gregb

This was actually a very easy problem, but just looks very intimidating, and takes some time.