Question

A closed curve encircles several conductors. The line integral AB around this curve is 3.83 * 10-4T # m. (a) 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.

Short answer

A) ${\mathrm{I}}_{\text{enet}}=305\mathrm{A}\text{b)}f\mathrm{B}.\mathrm{dl}=3.83\times {10}^{-4}\mathrm{Tm}$

Step-by-step solution

Concept of the net current in the conductor

We know that,$\mathit{f}\mathbf{\text{B.dl}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{I}}_{\mathbf{\text{enet}}}$ will give the net current in the conductor.

Calculate the net current in the conductor

Consider we have a closed curve encircles several conductors. The line integral $f\text{B.dl=}$around this curve is $3.83\times {10}^{-4}Tm$To find the net current in the conductors substitute the values in equation $\int B\cdot dl={\mu }_0{l}_{\text{enet}}$

$\begin{array}{r}3.83\times {10}^{-4}\mathrm{Tm}={\mu }_0{\mathrm{l}}_{\text{enet}}\\ {\mathrm{I}}_{\text{enet}}=\frac{3.83\times {10}^{-4}\mathrm{Tm}}{{\mu }_0}=\frac{3.83\times {10}^{-4}\mathrm{Tm}}{4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}}=305\mathrm{A}\end{array}$

If we instead went around the path in the opposite direction, the sign of the line integral would be reversed. Since at each point on the curve the direction of dl is reversed. That is$\text{fB.}\mathrm{dl}=3.83\times {10}^{-4}\mathrm{Tm}$