An Amperian loop encloses three conductors carrying currents of 4 A, 3 A, and 1 A. Two currents are directed into the page and one is directed out of the page. Which additional information is required to determine the enclosed current?
The direction of each current
Ampere’s law states \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\), where Ienc is the algebraic (net) sum of currents passing through the loop.
Currents in one direction are taken positive and in the opposite direction negative, so the sign of each current depends on its direction.
Knowing that two currents go in and one comes out is not enough; we must know which of the 4 A, 3 A and 1 A is directed out. For example, if 1 A is out, \(I_{enc} = 4 + 3 - 1 = 6\) A; if 4 A is out, \(I_{enc} = 3 + 1 - 4 = 0\).
The other data are not needed: resistance and material of the conductors do not enter Ampere’s law, and the loop radius does not change the enclosed current as long as the conductors are inside the loop.
Hence, the answer is The direction of each current.
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