Ampere's circuital law relates the line integral of the magnetic field around a closed path to:
The net current enclosed by the path.
Ampere's circuital law states \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\).
The line integral of B around any closed loop equals μ₀ times the net (algebraic) current passing through the surface bounded by the loop.
Resistance does not appear in the law; enclosed electric field is not part of it (circulation of E relates to changing flux in Faraday’s law); magnetic flux through a closed surface is the subject of Gauss’s law for magnetism, not Ampere’s law.
Hence, the answer is The net current enclosed by the path.
The magnetic field inside a long solenoid is 4 × 10−3 T when it carries a current of 2 A. If the current is increased to 6 A, the magnetic field becomes:
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?
A long solenoid carries a steady current. An Amperian loop is chosen completely outside the solenoid. What is the magnetic field along this loop according to the ideal solenoid model?
The wire loop shown in the figure carries a steady current $I$. Each straight section of the loop has length $d$. A part of the loop lies in xy plane and the other part is tilted at $30^\circ$ with respect to the xz plane. The magnitude of the magnetic dipole moment of the loop (in appropriate units) is

An infinitely long cylinder of radius $R$ carries a frozen-in magnetization $\vec M = ke^{-s^2} \hat z$, where $k$ is a constant and $s$ is the distance from the axis of cylinder. The magnetic permeability of free space is $\mu_o$. There is no free current present anywhere. The magnetic flux density ($\vec B$) inside the cylinder is