Three long, parallel conductors each carry a current of I = 2.02 A. The figure b

Question

Three long, parallel conductors each carry a current of I = 2.02 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 the following points.

(a) point A

_________ µT

out of the page?

into the page?

toward the left?

toward the right?

toward the top of the page?

toward the bottom of the page?

(b) point B

_________ µT

out of the page?

into the page?

toward the left?

toward the right?

toward the top of the page?

toward the bottom of the page?

(c) point C

________ µT

out of the page?

into the page?

toward the left?

toward the right?

toward the top of the page?

toward the bottom of the page?

_________ µT

out of the page?

into the page?

toward the left?

toward the right?

toward the top of the page?

toward the bottom of the page?

Explanation / Answer

At point B the direction of the magnetic fields due to top and bottom conductors will be in opposite direction and get cancelled. The effective field is only due to the third conductor

Bb = 2x10^-7 x2.02/(2x0.0095) = 21.26 uT

direction is towards bottom of the page

at point A, Magnetic fields due to the three conduct B due to the third conductor is shown

B3 = 2x10^-7 x2.02/(3x0.0095)=14.17 bottom of the page

B1 and B2

= 2x10^-7 x 2.02 /(sqrt(2)x0.95)x 10^-2) = 31.49 uT

These two fileds are at 90degrees

resulttant of B1 and B2 = 2x10^-7 x 2.02 /(0.95)x 10^-2)  =      42.52 uT towards page bottom aas shown.

Result B12 and B3 will add up as they are in the same direction

result = 42.52 +14.17 = 56.69 towards bottom of page

At Point C the field due to B1 and B2 will be same magnitude as that of A but opposite direction, towards page top

At C filed due to third conductor

B3 = 2x10^-7 x 2.02/(0.0095) = 42.52 uT towards bottom of the page

resultant of B1,B2 and B3 =42.52^2 - 42.52^2 = 0

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