Question

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are $I_1=+10A$, $I_2=-13A$, $I_3=+10A$, $I_4=+7A$, $I_5=-12A$ and $I_6=+18A$. The magnetic induction at a perpendicular distance of $10cm$ from the cable is $\left({\mu }_0=4\pi \times 10^{-7}Wb/A-m\right)$

• 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,
$i_{\text{net }}=i_1+i_2+i_3+i_4+i_5+i_6$
$i_{\text{net }}=10-13+10+7-12+18=20A$
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.
$=\frac{4\pi \times 10^{-7}\times 20}{2\pi \times 10\times 10^{-2}}=4\times 10^{-5}$
$B=40\mu T$