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, I4=−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
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a
5 μT
b
15 μT
c
74 μT
d
128 μT
answer is C.
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Detailed Solution
Net current is (20−6+12−7+18)A, i.e., 37 A.r=10100m=110mB=μ0I2πr=4π×10−7×37×102π×1T=74×10−6T=74 μT.