Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: `I_1=20A, I_2=-6A, I_3=12A, I_4=-7A, I_5=18A`. (Negative currents are opposite in direction to the positive.) The magnetic field induction at a distance of 10cm from the cable is (current enters at A and leaves at B and C as shown)

Options

A. `5 muT`
B. `15 muT`
C. `74 muT`
D. `128 muT`

Answer

C

Explanation

Net current is `(20-6+12-7+18)A i.e., 37A`
`r=10/100m=1/10m`
`B=(mu_(0)I)/(2pi)=(4pixx10^(-7)xx37xx10)/(2pixx1)T`
`=74xx10^(-6)T=74muT`