Three long, parallel conductors each carry a current of I = 2.20 A. The figure b
ID: 1391294 • Letter: T
Question
Three long, parallel conductors each carry a current of I = 2.20 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 0.75 cm, determine the magnitude and direction of the magnetic field at the following points.
(a) point A
2---Select---
to the left
to the right
toward the top of the page
toward the bottom of the page
into the page
out of the page
no direction
(b) point B
4---Select---to the left
to the right
toward the top of the page
toward the bottom of the page
into the page
out of the page
no direction
(c) point C
---Select---to the left
to the right
toward the top of the page
toward the bottom of the page
into the page
out of the page
no direction
magnitude ? ?T direction2---Select---
to the left
to the right
toward the top of the page
toward the bottom of the page
into the page
out of the page
no direction
Explanation / Answer
a)
Field is inversely proportional to the distance,
Field = = 2 *I*10^-7 / R = 2 * 2.2 x10^-7/ R = 4.4 * 10^-7 /R
At A.
If F1, F1 and F2 are the three fields.
Angle between the two F1 s is 90 degree and hence their resultant is
?2 * F1. And is directed from CtoA,
F1. = 4.4 * 10^-7 /R
F1. = 4.4 * 10^-7/?2 * a since R here is ?2*a
F2 = 4.4 * 10^-7/ 3a
?2 * F1 + F2 = 4.4 * 10^-7 (4/3) / a = 4.4*10^-7 (4/3) / 0.0075 = 7.82 * 10^-5 Wb
b)
At B
the field is due to F2 alone.
F2 = 4.4 * 10^-7/ (2a) = 4.4 * 10^-7/ (2*0.0075) = 2.93 * 10^-5 Wb
c)
At C
?2 * F1 - F2
4.4 * 10^-7 (2/3) / 0.0075)
3.91 * 10^-5 Wb from B to C
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