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Set 54 Problem number 3
A current of 4.3 Amps flowing clockwise in a coil consisting of 56 coplanar
loops, each of approximate radius .98 meters. Find the magnetic field at their common
center point of the loops (i.e., at the center of the coil).
The center point is at the same distance from every point of the loop.
Assume that the loop lies in the x-y plane.
- A clockwise current will result in every segment contributing a downward field
component, so all contributions reinforce one another.
- A line from the center to a point of the circle will be perpendicular to the
direction of the flow, so the displacement vector is perpendicular to the segment, and
sin(`theta) = 1.
- The total length of the loops is the circumference 2`pi r = 2`pi ( .98 m) = 351.6
m, multiplied by the number of loops 56, for a total of 19680 m.
- The field is B=k ' (IL)/r ^ 2 = .0000001 Tesla / Amp meter)( 84620 Amp m)/( .98 m) ^ 2 =
.00881 Tesla.
A loop of radius r can be thought of as a series of very short segments, of
total length 2 `pi r equal to the circumference of the circle.
- The segments can be made as short as desired, so the approximation to each field
can be made as accurate as desired.
- Each segment `dL is a source I `dL and is perpendicular to a line from the center
of the segment to the center of the circle.,
- Each segment thus contributes `dB = k ' I `dL / r^2 to the field at the center.
- All contributions are in the same direction, as can be easily verified.
When all the magnetic field contributions are added, we obtain
- B = `sum(`dB) = `sum( k ' I `dL / r^2 ).
Since k ' , I and r are identical for all contributions, we obtain
- B = `sum ( k ' I dL / r^2 ) = k ' I / r^2 * `sum(`dL).
Since the sum of all the `dL contributions for one loop is just the
circumference 2 `pi r of the loop, we finally have
- B = k ' I / r^2 * 2 `pi r = 2 `pi k ' I / r.
If there are N loops, then this result is multiplied by N.
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