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Funny differentiation

  • Thread starter bman!!
  • Start date
29
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we are given some component of a vector potential A

[tex]

A = B\displaystyle\int^d_0 I(t')\,dt'




[/tex]

where d = t -z/c
and B = constant=1/2 mu0 (permeability constant) x c (speed of light)


the derivation then wants me to calculate the magnetic field from this vector potential which is

[tex]
\vec B = \nabla \times A = \frac{d\(A}{dz} \vec j

[/tex]

where d/dz is meant to be partial diff

the result that is arrived at is

[tex] \vec B = CI(t-(z/c))\vec j[/tex]

where

C= +/- 1/2 x(permeability constant)


the step eludes me. i was fine until this point. i tried hitting it with the chain rule, and i suspect that the solution involves this somehow.

p.s. i'm new to latex, so if you think some information is missing, or the equations are simply not showing, It'd be great if someone could tell me, cheers.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

HallsofIvy
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Please clarify. You say that d= t- z/c but you haven't told us what that has to do with I(t').
 
tiny-tim
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… funny ? … it's the way you tell it … !

[tex]A = B\displaystyle\int^d_0 I(t')\,dt'[/tex]
Hi bman!! :smile:

You want ∂/∂z of [tex]\displaystyle\int^s_0 I(t')\,dt'[/tex] , where s depends on z.

(I'm using s instead of d, because d is ridiculous when you have derivatives. :frown:)

This is simply I(s) times (∂s/∂z).

Hint: Put I(t') = (d/dt')(J(t')). :smile:
 
29
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Hi bman!! :smile:

You want ∂/∂z of [tex]\displaystyle\int^s_0 I(t')\,dt'[/tex] , where s depends on z.

(I'm using s instead of d, because d is ridiculous when you have derivatives. :frown:)

This is simply I(s) times (∂s/∂z).

Hint: Put I(t') = (d/dt')(J(t')). :smile:
I'm assuming J(t') is the higher inegral function (from fundamental theorem of calculus:

[tex]\int_a^x f(t)dt= F(x) - F(a)[/tex]

where [tex] \frac{d}{dx}F(x) = f(x)[/tex]

i'm sorry for being really dense, but i always draw a blank with this type stuff.

so you eventually end up with ∂J/∂z = ∂s/∂z times ∂J(s)/∂z but ∂J(s)/∂s is just f(s), which is just I(t-z/c) from FTOC?

?

it seems so simple, but things like this just would never come to me in an exam.
 
tiny-tim
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… one step at a time … !

… so you eventually end up with ∂J/∂z = ∂s/∂z times ∂J(s)/∂z but ∂J(s)/∂s is just f(s), which is just I(t-z/c) from FTOC?

it seems so simple, but things like this just would never come to me in an exam.
Hi bman!! :smile:

Yeah … jus' the good ol' FTOC!

In an exam, remember that the examiners intend each question to be reasonably easy (unless there's only one question on the paper).

So just assume it's easy, and do it one step at a time!

In this case, you'd see the s (or d), and think "I obviously need to differentiate this … but how?", and then ask yourself what the s is there for. :smile:
 
29
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Hi bman!! :smile:

Yeah … jus' the good ol' FTOC!

In an exam, remember that the examiners intend each question to be reasonably easy (unless there's only one question on the paper).

So just assume it's easy, and do it one step at a time!

In this case, you'd see the s (or d), and think "I obviously need to differentiate this … but how?", and then ask yourself what the s is there for. :smile:
cheers mate, im just working through some electromagnetism, and stuff like this crops ups every half a page, where its simple to show, but still somewhat tricky, and the guy who wrote my notes, as clearly considered it trivial so he just doesnt bother showing the steps involved.

Im sure i'll have more questions soon enough ;)
 

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