Four long, parallel conductors carry equal 8 A currents. A cross-sectional view

Question

Four long, parallel conductors carry equal 8 A currents. A cross-sectional view of the conductors is shown in the figure. Each side of the square has length of 0.6 m. The current direction is out of the page at points indicated by the dots and into the page at points indicated by the crosses. Which of the diagrams correctly denotes the directions of the components of the magnetic field from each conductor at the point P? The permeability of free space is 4 pi times 10-7 N/A2 . What is the magnitude of each of the four components BA, BB, BC, and BD at the point What is the magnitude of the magnetic field at point P? Answer in units of T What is the direction of this resultant magnetic field?

Explanation / Answer

(part 1) By use of the right-hand rule, 4 is correct

(part 2) The magnitude of the field at point P is I/2r = ((4 x 10^-7)*8)/(2*.32) = 3.77 x 10^-6 T.

(part 3) The magnitude of the field is the vector sum of each of the individual fields, B = (3.77 x 10^-6)*4*sin45o = 1.07 x 10^-5 T.

(part 4) The field's direction is the vector sum of the individual fields, in the positive x-direction, so the answer is 5) +i.

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