

Biot Savart and Ampere's Laws
Name: Brian S.
Status: educator
Age: 20's
Location: N/A
Country: N/A
Date: 4/23/2004
Question:
I need some help explaining the difference
in the magnetic field calculated using
the Biot Savart Law and that for a solenoid
If you use the Biot Savart Law for a loop of wire
B = (mu)I/(2a) where a = radius of the loop, easy enough to prove if you
have n loops then B = n(mu)I/(2a) which would be similar to a solenoid, yet
now consider, a solenoid, the easiest way to find B
is Ampere's law
for a solenoid
B = (mu) I N where N = n/L loops per length
1) why do these equations differ?
2) if I use the solenoid eq for 1 loop.... n = 1 L = 2(pi) r
B = (mu) I/ 2(pi) r
This looks different than the equation for one loop of wire using the Biot
Savart Law.... B = (mu)I/(2a)
Replies:
Brian S.,
The Biot Savart law you use is for the center of a short coil. The length
of the coil must be much smaller than the radius. The formula is much more
complex when you go out from the loop, even if you stay on the loop's axis.
The solenoid formula is for a long coil. The length of the coil must be
much larger than the radius. The solenoid formula is the sum of the more
complex axis formulas for each coil. By assuming that length is so much
smaller than radius that (a/L) is tiny (essentially zero), the formula
simplifies to B=(mu)I(n/L). If you take the length to be essentially zero,
you get the Biot Savart law. Both are extremes of a much more complex
formula that I have never seen written out. For coils that are neither very
short or very long, the tradition is just to measure the field for a
standard current and scale it up or down as needed. The field will still be
linearly proportional to the current.
Dr. Ken Mellendorf
Physics Professor
Illinois Central College
These equations differ because they apply to different cases; either a
single current loop
or many current loops placed next to each other, N loops per unit length.
If you want to apply BiotSavart to a solenoid, remember that a given point
on the
axis of the solenoid is at the center of only one loop. For all other
loops, the B field
produced is less at that point, since each current element of a loop
produces a magnetic field element pointing at an angle to the axis of the
solenoid. Remember the vector dB
is perpendicular to both the current element and the vector from the current
element
to the point of observation. So a loop far away contributes almost nothing.
Your derivation using Ampere's Law has several bugs. First, Ampere's
law requires a large amount of symmetry to be simply applied. If you take,
for example, a path along the axis of the loop from  infinity to +
infinity,
the field varies in strength in a way you do not know; it is certainly NOT
constant.
Also when you calculated L, you took the circumference of the loop, but
you must take the length of the path; actually you need the line integral of
B.ds over a closed loop.
To use Ampere's Law for a solenoid, one normally assumes the solenoid is so
long
that the end effects are negligible in evaluating the line integral. Also
that, by
symmetry, the field is constant in strength and direction inside the
solenoid
(and again ignore the ends), and that the field is zero outside the
solenoid.
Then the calculation is "easy"!
Best, Dick Plano Professor Physics emeritus, Rutgers University
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Update: June 2012

