A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is (a) uniform in the region r a (b) zero in the region r a (c) directly proportional to r in the region r a (d) inversely proportional to r in the region r a

Answer: (c). For r a: Ienc = I, so B = (μ₀ I)/(2π r) ∝ 1/r. So the answer is (c).

A long straight wire of circular cross-section (radius a) is carrying… | JEE Main 2023 (6 Apr, Shift 1)

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Ampere's Law, Cylinders, Solenoid & Toroid

A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is

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Answer: (c)

For $r < a$ : $I_{enc} = I r^2/a^2$ , so $B = \dfrac{\mu_0 I r}{2\pi a^2} \propto r$ .
For $r > a$ : $I_{enc} = I$ , so $B = \dfrac{\mu_0 I}{2\pi r} \propto \dfrac{1}{r}$ .
So the answer is (c).

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