Four long, parallel conductors carry equal currents of I = 7.50 A. The figure be

Question

Four long, parallel conductors carry equal currents of I = 7.50 A. The figure below is an end view of the conductors. The current direction of the current is into the page at points A and B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square of edge length 0.200 m. Figure http://www.webassign.net/pse/p30-21.gif i get 3.01e-5 uT What is wrong with thise please help

Explanation / Answer

A=(-0.1, -0.1), B=(+0.1,-0.1), C=(-0.1,+0.1), D=(+0.1,+0.1), where 0.1=0.2/2 is half of square edge length; for a single wire B=I/(2pi*r), where r=0.1v2 is distance from any wire to point (0,0) which is the point in question. Thus vector B(A) = k*(i-j), vector B(B) = k*(i+j) vector B(C) = k*(i+j), vector B(D) = k*(i-j) where k= I/(2v2pi*r)= 7.5/(2v2pi*0.1v2)=1.59 net vector B= B(A) + B(B) + B(C) + B(D)= = 4*k*i = 47.7*i

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