Question

Six wires of current $I_1=1A,I_2=2A,I_3=3A,I_4=1A$ $I_5=4A$ and $I_6=5A$, cut the page perpendicularly at points $1,2,3,4,5$ and $5.$ The value of line integral of $\vec{B}$ around the dotted closed path (i.e. $\oint \vec{B}\cdot d\vec{l}$ ) is

• A. zero
• B. ${\mu }_{0}Wb{m}^{-1}$
• C. $2{\mu }_{0}Wb{m}^{-1}$
• D. $4{\mu }_{0}Wb{m}^{-1}$

Answer 1

Answer: (B)

$\oint \vec{B}\cdot d\vec{l}={\mu }_0\times$ Total current enclosed
$={\mu }_0\left[I_1+I_2+I_3+I_4+I_5\right)$
$={\mu }_0[1+2+3+(-1)+(-4)]$
$={\mu }_{0}Wb{m}^{-1}$