If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be _________

CONCEPT:

• Ampere’s Circuital Law states that the integrated magnetic field B along an imaginary closed path is equal to the multiplication of net electricity current enclosed inside this path and permeability of the medium.

\(\oint \overrightarrow{B}.\overrightarrow{dl} = μ _{0} I\)

where is B is the magnetic field μ₀ magnetic permeability, I is current enclosed, dl is the length.

EXPLANATION:

• Inside the conductor, Selecting any closed loop path inside the hollow conductor

current I = 0

From Amperes Circuital law

\(\smallint {\rm{B}}.{\rm{dl\;}} = {\rm{\;}}0\)

B = 0 (inside the pipe)

• Outside the conductor, selecting a closed path, there is current (in the pipe) passing inside this closed loop.

current I ≠ 0

\(\smallint {\rm{B}}.{\rm{dl\;}} \ne {\rm{\;}}0\)

• So the magnetic field associated with the hollow pipe current will be outside the pipe not inside the pipe.
• So the correct answer will be option 2.

• The magnetic field outside hollow pipe when this copper pipe carries a direct current, I

​\(B = \frac{{{μ _0}}}{{2\pi }}\frac{I}{r}\)