In a coaxial cable with current I on the inner conductor and −I on the outer conductor, what is the magnetic field in the region a < r < b (between the conductors)?

In a coaxial cable with current I on the inner conductor and −I on the outer conductor, what is the magnetic field in the region a < r < b (between the conductors)?

• A. B = μ0 I /(2π r)
• B. B = μ0 I r /(2π b)
• C. B = μ0 I /(π r)
• D. B = 0

Answer: (A)

The magnetic field around a long straight coaxial cable is circular about the axis, and its magnitude depends only on the radius r. Use Ampere’s law with a circular path of radius r in the region between the conductors (a < r < b). The current enclosed by this loop is just the inner conductor’s current I, since the outer conductor’s current −I lies at larger radius and isn’t enclosed by the loop. So ∮ B · dl = μ0 I_enc gives B(2π r) = μ0 I, hence B = μ0 I /(2π r). The field points azimuthally around the axis (direction given by the right-hand rule for the inner current). Outside the outer conductor (r > b), the enclosed current becomes I + (−I) = 0, so the field would be zero there. This is why the result in the region between conductors is proportional to 1/r rather than constant or growing with r.

The magnetic field around a long straight coaxial cable is circular about the axis, and its magnitude depends only on the radius r. Use Ampere’s law with a circular path of radius r in the region between the conductors (a < r < b). The current enclosed by this loop is just the inner conductor’s current I, since the outer conductor’s current −I lies at larger radius and isn’t enclosed by the loop.

So ∮ B · dl = μ0 I_enc gives B(2π r) = μ0 I, hence B = μ0 I /(2π r). The field points azimuthally around the axis (direction given by the right-hand rule for the inner current).

Outside the outer conductor (r > b), the enclosed current becomes I + (−I) = 0, so the field would be zero there. This is why the result in the region between conductors is proportional to 1/r rather than constant or growing with r.