Q 12-04-010NEETNEET 2021Top questionMedium
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 :
Answer: (D) see figure
Ampère's law with a circle of radius $r$:
Inside ($r < R$), the enclosed current is $I\dfrac{r^2}{R^2}$:
$$B \cdot 2\pi r = \mu_0 I\frac{r^2}{R^2} \;\Rightarrow\; B = \frac{\mu_0 I r}{2\pi R^2} \propto r$$
Outside ($r > R$):
$$B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$$
$B$ rises linearly from zero to a maximum at $r = R$, then falls as $1/r$. This is option (4).
Solution by Sreeraj P, M.Sc Physics