Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

From Ampere's circuital law, for a long straight wire of circular cross-section carrying a steady current, the variation of the magnetic field inside and outside the region of the wire is:

A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region.

Uniform and remains constant for both regions.

A linearly increasing function of distance upto the boundary of the wire and then a linearly decreasing one for the outside region.

A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1/r dependence for the outside region.

Step-by-Step Solution

Inside the Wire ( $r < a$ ): Assuming the current $I$ is uniformly distributed over the cross-section of radius $a$ , the current enclosed by a loop of radius $r$ is $I_{enclosed} = I(r^2/a^2)$ . Applying Ampere's Law: $B(2\pi r) = \mu_0 I(r^2/a^2) \implies B = \left(\frac{\mu_0 I}{2\pi a^2}\right)r$ . Thus, the magnetic field is directly proportional to the distance ( $B \propto r$ ), representing a linear increase.
Outside the Wire ( $r > a$ ): The current enclosed is the total current $I$ . Applying Ampere's Law: $B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$ . Thus, the magnetic field is inversely proportional to the distance ( $B \propto 1/r$ ).
Conclusion: The field increases linearly up to the boundary and then decreases with $1/r$ dependence outside.

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