Question

why can't i use ampere's law here
http://assets.openstudy.com/updates/attachments/56a0e32ce4b0feea8a8b605f-ganeshie8-1453384502777-zz.png

Answer 1

Using biot savart law and integrating over L gives me B = 50 nT

http://www.wolframalpha.com/input/?i=2*%5Cint_0%5E%280.18%2F2%29+%281.26*10%5E%28-6%29%2F%284*pi%29*0.0582*0.131%2F%28s%5E2%2B0.131%5E2%29%5E%283%2F2%29%29+ds


is there any other alternative way?

Answer 2

sorry thats beyond what i know

Answer 3

https://en.wikiversity.org/wiki/Biot-Savart_Law

the derivation is for an infinite wire - which is why the question contains a warning, i guess

Answer 4

why can't i use ampere's law here

Is a very interesting question.
You have a nice uniform B-field to do your B.dl integration and you know exactly how much current is going through the area defined by your Amperian loop.

I suspect it is something to do with the fact that you cannot realise such a situation in isolation.
You need to get the current in at one end and out at the other.
OK you have a \[\LARGE \bigsqcup\]shaped arrangement with the bottom of your U being the bit of conductor whose B-field you want to find.
Unfortunately the other two bits of wire carrying the current in and out will also produce a B-field and there is no Amperian loop where the B-field is constant, so the integration cannot easily be done.

OK, what about a single straight piece of wire with charges moving along (a current) and accumulating at the ends.
Now there are no extra bits of current carrying conductor producing extra B-field.
Unfortunately this accumulation of charge will produce an E-field which changes with time passing through the Amperian loop - a displacement current which has to be dealt with on the mu I side of the equation.

I have no idea whether or not these are valid reasons for not being able to use Ampere's Law???

Answer 5

the problem can be represented by this drawing:

Answer 6

where the horizontal axis is the \(x-\)axis, whereas the vertical axis, is the \(z-\) axis. Now, we can write:
\[ \Large d{\mathbf{B}} = \frac{i}{{c{r^3}}}\left( {d{\mathbf{l}} \times {\mathbf{r}}} \right)\]
and developing the computation with such reference system, we get:
\[\Large \begin{gathered}
d{\mathbf{B}} = \frac{i}{{c{r^3}}}\left( {d{\mathbf{l}} \times {\mathbf{r}}} \right) = \hfill \\
\hfill \\
= \frac{i}{{c{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}\det \left( {\begin{array}{*{20}{c}}
{{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\
0&0&{dz} \\
R&0&z
\end{array}} \right) = \hfill \\
\hfill \\
= \frac{{iR}}{{c{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}dz\;{\mathbf{\hat y}} \hfill \\
\end{gathered} \]
Next, after a simple integration, we can write:
\[\Large {\mathbf{B}} = \frac{{iR}}{c}\int_{ - L/2}^{L/2} {\frac{{dz}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}}} \;{\mathbf{\hat y}} = \frac{i}{{cR}}\frac{L}{{\sqrt {{R^2} + \frac{{{L^2}}}{4}} }}\;{\mathbf{\hat y}}\]
where the subsequent identity, can be useful:
\[\Large \int {\frac{{dz}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}} = \frac{z}{{\sqrt {{z^2} + {R^2}} }}} \]
Finally, we can note this limit relation:
\[\Large \mathop {\lim }\limits_{L \to + \infty } {\mathbf{B}} = \mathop {\lim }\limits_{L \to + \infty } \frac{i}{{cR}}\frac{L}{{\sqrt {{R^2} + \frac{{{L^2}}}{4}} }}\;{\mathbf{\hat y}} = \frac{{2i}}{{cR}}\;{\mathbf{\hat y}}\]
which is the \(Biot-Savart\) law, according to the observation of @IrishBoy123

Answer 7

oops.. I have made a typo:
\[\Large \boxed{\int {\frac{{dz}}{{{{\left( {{z^2} + {R^2}} \right)}^{3/2}}}} = \frac{1}{{{R^2}}}\frac{z}{{\sqrt {{z^2} + {R^2}} }}} }\]
and, as usually, I have used the \(CGS\) system

Answer 8

The Biot-Savart method is the standard way to solve this problem but why not Ampere's law?

Answer 9

Thank you all :)
I guess the finite wire segment has no cylindrical symmetry as mentioned by @Farcher /@IrishBoy123 . To use ampere's law we must have B to be same along the entire cylindrical surface...

since ampere's law works only for infinite wires or points that are in middle and near to the wire, we cannot use ampere's law here

Does that sound good ?

@Michele_Laino @Farcher @IrishBoy123