Question

the figure shows three wires, each carrying a current of 0.406 a. what is the absolute value of $oint vec{b} cdot d vec{s}$ for loop a?

Explanation:

Step1: Apply Ampere's Law

Ampere's Law is given by $\oint \vec{B}\cdot d\vec{s}=\mu_0i_{enc}$. Here, we need to find the enclosed current $i_{enc}$ for loop $a$.

Step2: Determine the enclosed current

Looking at loop $a$, the current going into the page (using the right - hand rule, we consider the direction). The top wire has current in one direction (let's assume for Ampere's Law sign convention). The enclosed current $i_{enc}=- 0.406A$ (negative because of the direction, but we are asked for the absolute value).

Answer:

$| \oint \vec{B}\cdot d\vec{s}|=\mu_0|i_{enc}|$. Since $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$ and $|i_{enc}| = 0.406A$, $| \oint \vec{B}\cdot d\vec{s}|=4\pi\times10^{-7}\times0.406\approx5.1\times 10^{-7}\ T\cdot m$