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 positve.) 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. `5muT`
B. `15muT`
C. `74muT`
D. `128muT`

Answer

C

Explanation

(c) Net current is `(20-6+12-7+18)A`, i.e., 37A.
`r=10/100m=1/10m`
`B=(mu_0I)/(2pir)=(4pixx10^-7xx37xx10)/(2pixx1)T=74xx10^-6T`
`=74muT.`