Chapter 4: Q41E (page 950)

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

Expert verified

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

Step by step solution

Concept of the net current in the conductor

We know that, $f B.dl = μ_{0} I_{enet}$ 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 $f B.dl=$ around this curve is $3.83 \times 10^{- 4} T m$ To find the net current in the conductors substitute the values in equation $\int B \cdot d l = μ_{0} l_{enet}$

$\begin{aligned} 3.83 \times 10^{- 4} Tm = μ_{0} l_{enet} \\ I_{enet} = \frac{3.83 \times 10^{- 4} Tm}{μ_{0}} = \frac{3.83 \times 10^{- 4} Tm}{4 π \times 10^{- 7} Tm / A} = 305 A \end{aligned}$

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 $fB. dl = 3.83 \times 10^{- 4} Tm$