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Set 54 Problem number 3
A coil consists of 140 coplanar loops, each of approximate radius .2 meters.
Find the magnetic field at their common center point if a current of 1.3 Amps flows
clockwise in 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 ( .2 m) = 879.2
m, multiplied by the number of loops 140, for a total of 123000 m.
- The field is B=k ' (IL)/r ^ 2 = .0000001 Tesla / Amp meter)( 159900 Amp m)/( .2 m) ^ 2 =
.3997 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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