Question

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are $I_1 = + 10 \,A$, $I_2 = - 13\, A$, $I_3 = + 10\, A$, $I_4 = + 7\,A$, $I_5 = - 12\, A$ and $I_6 = + 18 \,A$. The magnetic induction at a perpendicular distance of $10 \,cm$ from the cable is $\left(\mu_{0}=4\pi\times10^{-7}\,Wb/A-m\right)$

• A. $40\,\mu T$
• B. $37.5\,\mu T$
• C. $30\,\mu T$
• D. $35\,\mu T$

Answer 1

Answer: (A)

Net current due to all wires,
$i_{\text {net }}=i_{1}+i_{2}+i_{3}+i_{4}+i_{5}+i_{6}$
$i_{\text {net }}=10-13+10+7-12+18=20 A$
We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance $r$ from it is given by
$B=\frac{\mu_{0} i}{2 \pi r}=\frac{\mu_{0} i_{\text {net }}}{2 \pi r}$
where, $i=$ current in wire
and $r=$ perpendicular distance.
$=\frac{4 \pi \times 10^{-7} \times 20}{2 \pi \times 10 \times 10^{-2}}=4 \times 10^{-5}$
$B =40 \mu T$