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    				| 1. | 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)A. `5muT`B. `15muT`C. `74muT`D. `128muT` | 
| Answer» Correct Answer - C (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.` | |