A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by :

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

⇒ B_inside ∝ r

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

 ⇒ B_outside ∝ \(\frac{1}{r}\)

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

Hence, option 3) is the correct answer.