Question

Six wires of current $I_{1}=1 A , I_{2}=2 A , I_{3}=3 A , I_{4}=1 A$ $I_{5}=4 A$ and $I_{6}=5 A$, 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
$=\mu_{0}\left[I_{1}+I_{2}+I_{3}+I_{4}+I_{5}\right)$
$=\mu_{0}[1+2+3+(-1)+(-4)]$
$=\mu_{0} \,Wb \,m ^{-1}$