A straight current segment of length .005 m carries a current of 1.9 Amps. A field detector lies 2 meters from the segment. The vector from the midpoint of the segment to the detector makes at angle 78 degrees with the segment. What is the strength of the field measured by the detector?
Since the length of the segment is much less than the distance, we are justified in approximating the field by considering the source I * L to lie at a nearly uniform distance from the point at which the field is calculated.
Because the angle between the direction of I * L and the vector from source to field point is not 90 degrees, the field strength is less than k ' ( I * L ) / r^2. As the angle changes from 90 degrees to 0, the field strength decreases from this maximum value to 0 according to the sine function, with field strength B = k ' [(IL)/r ^ 2](sin `theta).
If the vector was perpendicular the angle would be 90 deg, the sine would be 1, and the field would be just k ' IL/r ^ 2 = 2.375E-10 Tesla; the given angle reduces this by factor sin( 78 deg) = .9778, giving the answer specified above.
The magnitude of a magnetic field due to source I*L , at distance r from the source such that the angle between I and a line from the source to the point is `theta, is B = k ' (IL) / r^2 * sin(`theta).
Since sin(`theta) varies from 0 to 1 as `theta varies from 0 to 90 deg, the field has maximum strength k ' IL / r^2 when `theta = 90 deg, and is reduced for any other angle, reaching 0 when `theta = 0.
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