A thick current-carrying cable of radius 'R' carries current 'I' uniformly distributed across its cross-sectio

Question

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:

Accepted Answer

1.  **Region 1: Inside the cable ($r < R$):**
    Apply Ampere's Circuital Law: $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enclosed}$.
    For a path of radius $r$ inside the wire, the current enclosed is a fraction of the total current $I$, proportional to the area:
    $$I_{enclosed} = I \left(\frac{\pi r^2}{\pi R^2}\right) = I \frac{r^2}{R^2}$$
    $$B(2\pi r) = \mu_0 I \frac{r^2}{R^2}$$
    $$B = \left(\frac{\mu_0 I}{2\pi R^2}\right) r$$
    Thus, $B \propto r$ (Linear increase passing through the origin).
2.  **Region 2: Outside the cable ($r \geq R$):**
    The enclosed current is the total current $I$.
    $$B(2\pi r) = \mu_0 I$$
    $$B = \frac{\mu_0 I}{2\pi r}$$
    Thus, $B \propto \frac{1}{r}$ (Rectangular hyperbola decrease).
3.  **Conclusion:** The magnetic field increases linearly from zero at the axis to a maximum at the surface ($r=R$), and then decreases inversely with distance outside. This corresponds to the standard graph for a solid cylindrical wire .

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