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NEET 2026 · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from NEET 2026

Q3NEET 2026Easy

Ampere's Law, Cylinders, Solenoid & Toroid

The figure given below shows a long straight solid wire of circular cross-section of radius ' aa ' carrying steady current II. The current II is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B)(B) with distance (r)(r) from the axis of the conductor in the region is:

Figure

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Answer: (a)

  1. Inside (r<ar < a): Ienc=Ir2a2I_{enc} = I\frac{r^2}{a^2}, so B=μ0Ir2πa2∝rB = \frac{\mu_0 I r}{2\pi a^2} \propto r (straight line from origin).
  2. Outside (r>ar > a): B=μ0I2πr∝1rB = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}, maximum at r=ar = a.
  3. So the answer is (a): linear rise to r=ar = a, then 1/r1/r fall.
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