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Erik Strand
pit
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ebdf24e3
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ebdf24e3
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6 years ago
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Erik Strand
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Answer 6.4
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@@ -189,18 +189,49 @@ Yes, there is a sign error that I'm ignoring for now.
...
@@ -189,18 +189,49 @@ Yes, there is a sign error that I'm ignoring for now.
The ampere was formerly defined [BIPM, 2014] as "The ampere is that constant current which, if
The ampere was formerly defined [BIPM, 2014] as "The ampere is that constant current which, if
maintained in two straight parallel conductors of infinite length, of negligible circular
maintained in two straight parallel conductors of infinite length, of negligible circular
cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force
cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force
equal to
2 × 10−7
newton per metre of length."
equal to
$$
\n
um{2e-7}$$
newton per metre of length."
### (a)
### (a)
{:.question}
{:.question}
Show that that current at that distance produces that force.
Show that that current at that distance produces that force.
First let's find the magnetic field of an infinitely long straight conductor. Considering the
Biot-Savart Law and the symmetry of this problem, the magnetic field must have no component parallel
to the wire, must always be perpendicular to the radial separation between the test point and the
wire, and must have a constant magnitude at each radius. Consider then a circle of radius $$r$$
centered on the wire. Amp
è
re's Law tells us that the magnitude of the field on this circle is
$$I / (2
\p
i r)$$.
The differential force exerted by this field on a differential piece of current is $$dF = I (dl
\t
imes B)$$. In this case the direction of the current and the magnetic field are perpendicular, so
the direction of the cross product is always toward the other wire. The magnitude of the force per
meter is
$$
\b
egin{align
*
}
F &=
\f
rac{
\m
u_0 I^2}{2
\p
i r}
\\
&=
\f
rac{4
\p
i
\t
imes 10^{-7}
\s
i{H/m}
\c
dot 1
\s
i{A^2}}{2
\p
i
\c
dot 1
\s
i{m}}
\\
&=
\n
um{2e-7}
\s
i{N}
\e
nd{align
*
}
$$
Note that we don't need to multiply this by two to account for the force wire two exerts on wire
one. Newton's
[
third
law
](
https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_third_law
)
guarantees that
this is equal and opposite, so it only makes sense to talk about the force exerted by one on the
other. Maxwell's equations respect this (otherwise it would be easy to build a perpetual motion
machine).
### (b)
### (b)
{:.question}
{:.question}
What is the problem with defining the ampere this way?
What is the problem with defining the ampere this way?
It's not a very practical experiment. No wires are infinitely long, so you'll necessarily have
fringe fields to factor into your results. No conductors are infinitely thin. You have to hold the
wires and supply current to them somehow, so they can't be totally surrounded by a vacuum.
## (6.5)
## (6.5)
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