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 :
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Answer
(4) [figure]
Solution
Using Ampere's circuital law with a circle of radius r coaxial with the cable:
Inside (r < R): current enclosed = I r2R2
⇒ B(2 r) = 0 I r2R2 ⇒ B = 0 I r2 R2 ⇒ B r (straight line through origin)
Outside (r ≥ R): current enclosed = I
⇒ B = 0 I2 r ⇒ B 1r (decreases)
B is maximum at the surface (r = R).
∴ graph (4) is correct.
Answer: (4)