Suppose that a long hollow rod has current flowing through it along the length of the rod. I am trying to solve a problem which has such an arrangement of current. The problem manual states that the magnetic field lines will be the same as if the current were arranged uniformly in the rod (not hollow). I can see that the rod with uniform current can be divided into differential rings with uniform current. In that case, would the magnetic field be a superposition of the magnetic field of all such rings?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Are you sure that you have quoted the problem correctly? It is true that externally the magnetic field is the same as that of a uniform conductor, but internally the fields differ. The equivalence of the external field can easily be seen from an application of Ampere's law to a loop around the outside of the conductor. Indeed this is also a good way to derive the internal field, using a loop within the conductor.
On your specific question: yes the magnetic field is a linear function of the current distribution, so we can superpose the fields due to any current elements.
-
$\begingroup$ I can see that the RHS of Ampere’s law will be the same as the case of the uniform current, but how did you solve the line integral to find out the fields were equal? Also, wouldn’t the internal field be zero since there is no enclosed current? $\endgroup$ Jul 30, 2022 at 21:04
-
$\begingroup$ Hint: use a circular amperian loop, Use Cylindrical symmetry to Extract $\vec{B}$ out of the integral $\int \vec{B} \cdot \vec{dl}$ remember it must be an infinite rod to use amperes law without the maxwell displacement current. $\endgroup$ Jul 30, 2022 at 22:31
-
$\begingroup$ Yes @Andromeda, the field in the hollow space is zero. That's why I said that the internal field is different from the case of a solid cylindrical conductor $\endgroup$– CWPPJul 31, 2022 at 5:50