A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be :

CONCEPT:

According to the Ampere's circuital law,

The magnetic field inside the current carrying cable is written as:

\(B = \frac{{{\mu _0}I}}{{2\pi {R^2}}}.r\)

and the magnetic field outside the current carrying cable is written as:

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

Here, B is the magnetic field, " I " is the current, and r is the distance.

CALCULATION:

From Ampere's circuital law

\(B = \frac{{{\mu _0}I}}{{2\pi {R^2}}}.r\) if r < R

⇒ Binside ∝ r      ----- (1)

\(B = \frac{{{\mu _0}I}}{{2\pi r}}\) if r ≥  R

 ⇒ Boutside ∝ \(\frac{1}{r}\)       ----- (2)

Hence, magnetic field B with distance r from the axis of the cable is given as

∴ From equation (1) for r < R, Binside ∝ r

Hence,  option 2) is the correct choice.