Consider we have a closed encircles several conductors. The line integrtal \(\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}\) around this curve is \(\displaystyle{3.83}\times{10}^{{-{4}}}{T}\cdot{m}\)

a) We need to find the net current in the conductors. We know that,

\(\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}=\mu_{{0}}{I}_{{{e}{n}{c}{l}}}\)

so,

\(\displaystyle{3.83}\times{10}^{{-{4}}}{T}\cdot{m}=\mu_{{0}}{I}_{{{e}{n}{c}{l}}}\)

or, \(\displaystyle{I}_{{{e}{n}{c}{l}}}={\frac{{{3.83}\times{10}^{{-{4}}}{T}\cdot{m}}}{{\mu_{{0}}}}}={\frac{{{3.83}\times{10}^{{-{4}}}{T}\cdot{m}}}{{{4}\pi\times{10}^{{-{7}}}{T}\cdot\frac{{m}}{{A}}}}}={305}{A}\)

If we instead went around the path in the opposite direction, the sign of the line integral would be reserved. Since at each point on the curve the direction of \(\displaystyle{d}\vec{{{l}}}\) is reserved. That is,

\(\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}=-{3.83}\times{10}^{{-{4}}}{T}\cdot{m}\)

a) We need to find the net current in the conductors. We know that,

\(\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}=\mu_{{0}}{I}_{{{e}{n}{c}{l}}}\)

so,

\(\displaystyle{3.83}\times{10}^{{-{4}}}{T}\cdot{m}=\mu_{{0}}{I}_{{{e}{n}{c}{l}}}\)

or, \(\displaystyle{I}_{{{e}{n}{c}{l}}}={\frac{{{3.83}\times{10}^{{-{4}}}{T}\cdot{m}}}{{\mu_{{0}}}}}={\frac{{{3.83}\times{10}^{{-{4}}}{T}\cdot{m}}}{{{4}\pi\times{10}^{{-{7}}}{T}\cdot\frac{{m}}{{A}}}}}={305}{A}\)

If we instead went around the path in the opposite direction, the sign of the line integral would be reserved. Since at each point on the curve the direction of \(\displaystyle{d}\vec{{{l}}}\) is reserved. That is,

\(\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}=-{3.83}\times{10}^{{-{4}}}{T}\cdot{m}\)