Ampere's law in integral form for steady currents?

Ampere's law in integral form for steady currents?

• A. ∮ B · dl = μ0 ε0 dΦ_E/dt
• B. ∮ E · dl = 0
• C. ∮ B · dl = μ0 I_enc
• D. ∮ B · dl = - dΦ_B/dt

Answer: (C)

Circulation of the magnetic field around a closed loop is proportional to the current threading that loop. For steady currents, the displacement current term vanishes, so the integral of B around a closed path equals μ0 times the current enclosed by that path. This is Ampere's law in integral form for steady currents. The other forms correspond to different situations: the line integral of E around a loop is zero in electrostatics, Faraday’s law relates ∮ B to the time rate of change of electric flux in the context of changing magnetic fields around a loop, and the full Maxwell-Ampere law includes the displacement current term ε0 dΦ_E/dt, which is not needed when currents are steady.

Circulation of the magnetic field around a closed loop is proportional to the current threading that loop. For steady currents, the displacement current term vanishes, so the integral of B around a closed path equals μ0 times the current enclosed by that path. This is Ampere's law in integral form for steady currents.

The other forms correspond to different situations: the line integral of E around a loop is zero in electrostatics, Faraday’s law relates ∮ B to the time rate of change of electric flux in the context of changing magnetic fields around a loop, and the full Maxwell-Ampere law includes the displacement current term ε0 dΦ_E/dt, which is not needed when currents are steady.