Question

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)

• A. $5\mu T$
• B. $15\mu T$
• C. $74\mu T$
• D. $128\mu T$

Answer 1

Answer: (C)

$74\mu T$

Net current is $(20-6+12-7+18)A=37A$

$r=\frac{10}{100}m=\frac{1}{10}m$

$B=\frac{{\mu }_0}{2\pi r}=\frac{4\pi \times {10}^{-7}\times 37\times 10}{2\pi +1}T=74\times {10}^{-6}T$$=74\mu T$