Three long, parallel conductors carry currents of I = 2.48 A. The figure below i

Question

Three long, parallel conductors carry currents of I = 2.48 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 0.95 cm, determine the magnitude and direction of the magnetic field at points A, B, and C.

BA : Magnitude   
. Three long, parallel conductors carry currents of I = 2.48 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 0.95 cm, determine the magnitude and direction of the magnetic field at points A, B, and C.

Explanation / Answer

Field is inversely proportional to the distance,
Field = = 2Ix10^-7/ R = 2*2.48 x10^-7/ R = 4.96e-7/R

At A.
If F1, F1 and F2 are the three fields.
Angle between the two F1 s is 90 ? and hence their resultant is
sqrt 2 * F1. And is directed from CtoA,
F1. = 4.96e-7/R
F1. = 4.96e-7/sqrt 2*a since R here is sqrt2*a
F2 = 4.96e-7/ 3a
sqrt2 * F1 + F2 = 4.96e-7 {(4/3) / a = 4.96e-7 {(4/3) / 0.0095 = 6.96x10^-5 Wb
======================================...

At B
the field is due to F2 alone.
F2 = 4.96e-7/ (2a) = 4.96e-7/ (2*0.0095) = 2.61x10^-5 Wb
=====================================
At C
?2 * F1 - F2
4.97e-7 {(2/3)/0.0095)
3.48x10^--5 Wb from B to C
======================================...

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