question 6 6. Four long parallel conductors carry equal currents of I- 5.00A. Th

Question

question 6

6. Four long parallel conductors carry equal currents of I- 5.00A. The figure below is a cross-sectional view of the conductors. The current directions are indicated in the figure. (a) Calculate the magnitude and direction of the magnetic field at point C. (b) Calculate the magnitude and direction of the force per unit length on the conductor located at point C. The length of each side of the square is 0.200 m. CAns: a) 7.9 x 10"T at 71.6° below the tx- aris, b 3.95x10 N at 18.4 below the x-axis)

Explanation / Answer

(A) due to A ,

Ba = u0 I / (2 pi r)

Ba = (4pi x 10^-7 x 5 / (2 x pi x 0.2)) (-j)

= - 5 x 10^-5 j^

due to B,

Bb = (4 x pi x 10^-7 x 5 / (2 x pi x sqrt(0.2^2 + 0.2^2))) (cos45i - sin45j)

Bb = 2.5 x 10^-6 i^ - 2.5 x 10^-6 j^

due to D,

Bd =- 5 x 10^-6 i

Bnet = Ba + Bb + Bd = -2.5 x 10^-6i^ - 7.5 x 10^-6 j^

magnitude = sqrt(2.5^2 + 7.5^2) x 10^-6

= 7.9 x 10^-6 T

direction = tan^-1(7.5 /2.5) = 71.6 deg below the -x axis.

(B) F = IL X B

= (5A)k x -2.5 x 10^-6i^ - 7.5 x 10^-6 j^

= - 12.5 x 10^-6 j^ + 37.5 x 10^-6 i^

magnitude = sqrt(12.5^2 + 37.5^2)x 10^-6

= 3.95 x 10^-5 N

direction = tan^-1(12.5 / 37.5) = 18.4 deg below the +x axis

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