Question

each of the eight conductors in the figure carries 2.36 a of current into or out of the page. two paths are indicated for the line integral ∮b⃗ ⋅ds⃗. what is the value of the integral for (a) path 1 and (b) path 2? (a) number units t·m (b) number units t·m did you use a right - hand rule for amperes law to determine the sign of the current encircled by an amperian loop? did you include all encircled currents, along with their signs? hint etextbook and media assistance used

Explanation:

Step1: Recall Ampere's law

Ampere's law is given by $\oint\vec{B}\cdot d\vec{s}=\mu_0I_{enclosed}$, where $\mu_0 = 4\pi\times10^{- 7}\ T\cdot m/A$ and $I_{enclosed}$ is the net current enclosed by the Amperian loop.

Step2: Determine the current enclosed for path (a)

For path (a), assume that the current direction is determined using the right - hand rule. If we consider the direction of the loop and the current directions of the conductors, and assume that the current into the page is negative and out of the page is positive. If all four conductors have current $I = 2.36\ A$ and are either all into or all out of the page, and the loop encloses 4 conductors. So $I_{enclosed,a}=4\times2.36\ A = 9.44\ A$. Then $\oint\vec{B}\cdot d\vec{s}=\mu_0I_{enclosed,a}=(4\pi\times10^{-7}\ T\cdot m/A)\times9.44\ A\approx1.19\times10^{-5}\ T\cdot m$.

Step3: Determine the current enclosed for path (b)

For path (b), assume the same current direction convention. If the loop encloses 0 conductors, then $I_{enclosed,b}=0\ A$. So $\oint\vec{B}\cdot d\vec{s}=\mu_0I_{enclosed,b}=0\ T\cdot m$.

Answer:

(a) $1.19\times 10^{-5}$; Units: T·m
(b) 0; Units: T·m