# A closed curve encircles several conductors. The line integral \oint\vec{B}\c

A closed curve encircles several conductors. The line integral $$\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}$$ around this curve is $$\displaystyle{3.83}\times{10}^{{-{4}}}{T}\cdot{m}$$ What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Troy Lesure
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,
$$\displaystyle\oint\vec{{{B}}}\cdot{d}\vec{{{l}}}=-{3.83}\times{10}^{{-{4}}}{T}\cdot{m}$$