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Q.

Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: I1 = 20 A, I2 = -6 A, I3 = 12 A, 14 = -7A, I5= 18 A. (Negative currents are opposite in direction to the positive.) The magnetic field induction at a distance of 10 cm from the cable is (current enters at A and leaves at B and C as shown)

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a

5mT

b

15 mT

c

74mT

d

128mT

answer is C.

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Detailed Solution

Net current is (20 - 6 + 12 - 7 + 18) A, i.e.,37 A.r=10100 m=110 m B=μ0I2πr=4π×10-7×37×102π×1 T=74×10-6 T=74μT

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Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: I1 = 20 A, I2 = -6 A, I3 = 12 A, 14 = -7A, I5= 18 A. (Negative currents are opposite in direction to the positive.) The magnetic field induction at a distance of 10 cm from the cable is (current enters at A and leaves at B and C as shown)