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According to Ampere's Circuital Law, the line integral \[\oint\vec{B}\cdot d\vec{l}\] taken around any closed loop equals:

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Question

According to Ampere's Circuital Law, the line integral \[\oint\vec{B}\cdot d\vec{l}\] taken around any closed loop equals:

Options

  • \[\mu_0\] times the net steady current passing through the loop

  • \[\mu_0\] times the net current divided by the loop's perimeter

  • The net current enclosed divided by \[\mu_0\]

  • Zero for every closed loop

MCQ
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Solution

The law states that \[\oint\vec{B}\cdot d\vec{l}=\mu_0 I\], where \[I\] is the net steady current passing through the loop. The result holds for any closed Amperian loop, not just circular ones.

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