Homework Answers
Using Ampere's law, for closed loop:
\(\oint B \cdot d s=\mu_{0} * I_{n e t}\)
\(\mu_{0}=\) Magnetic constant \(=4^{*} \mathrm{pi}^{*} 10^{\wedge}-7\)
\(I_{\text {net }}=\) Net current in the window frame \(=3.0 \mathrm{~A}\)
So,
\(\oint B . d s=\mu_{0} * I_{n e t}=4 \pi * 10^{-7} * 3.0=3.8 * 10^{-6} \mathrm{~T} . \mathrm{m}\)
Correct option is \(\mathrm{D}\).
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