Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

Graph showing B linearly increasing for r < R and decreasing as 1/r for r > R

Graph showing B constant for r < R and decreasing for r > R

Graph showing B zero for r < R and decreasing for r > R

Graph showing B increasing parabolically for r < R and decreasing for r > R

Step-by-Step Solution

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).
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).
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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