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Well, this popped inside my head when I was doing Boit-Savart's law.

$$d\vec{B}=\frac{\mu_{0}I d\vec{l} \times \vec{r}}{r^3}$$

l is the vector that represents the current element (i.e the direction of current flow) and r represents the point at which we have to find the magnetic field. So from this can we infer that the Magnetic field inside a conductor is uniform as the $\vec{l}$ and $\vec{r}$ are in the same direction and thus the cross product is 0. So dB=0 and thus B is uniform along the vector that we have taken...

Elaborated: Take a point P inside the conductor. Now take another point Q such that $\vec{PQ}$ is parallel to the surface. Let $P$ and $Q$ be at a distance $a$ from the center. Then $\vec{r}$ and $\vec{l}$ are in the same line $\vec{PQ}$ making the angle between them zero. So $dB=0$. Therefore $B$ remains uniform along the vector $\vec{PQ}$.

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    $\begingroup$ Conductor has non-zero thickness, so $\vec{l}$ really is $\vec{j} dV$ and $\vec{j}$ is not of same direction as $\vec{r}$. $\endgroup$ Mar 4, 2023 at 17:48
  • $\begingroup$ If dB is zero for some current element this means just that the contribution of this current element to the field at point Q is zero. But the field at Q is given by the integral of all contributions, from all current elements in the wire. $\endgroup$
    – nasu
    Mar 7, 2023 at 13:16

1 Answer 1

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For DC current the field increases from zero on the axis of the conductor to a maximum value on the surface. It is not uniform across the cross section of the conductor. Of course, it is constant in the sense that there is no time dependence (what "constant" usually means).

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  • $\begingroup$ it means same value along the vector that we taking inside the conductor... $\endgroup$
    – Sanjay
    Mar 4, 2023 at 17:22
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    $\begingroup$ No, this is usually called uniform field. See for example the electric field inside a charged capacitor. It is "uniform field" and not constant field. Constant referes to the time dependence. $\endgroup$
    – nasu
    Mar 5, 2023 at 0:27
  • $\begingroup$ Thanks for saying that. I have changed the question accordingly... Can you comment the answer now?? $\endgroup$
    – Sanjay
    Mar 7, 2023 at 5:54
  • $\begingroup$ There is no other answer to comment anymore. $\endgroup$
    – nasu
    Mar 7, 2023 at 13:17

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