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JEE Main 2022 (25 Jun, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2022 (25 Jun, Shift 1)

Q32JEE Main 2022 · 25 JUN · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be:

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Answer: (b)

  1. For r<Rr < R: B(2πr)=μ0Ir2R2B (2\pi r) = \mu_0 I \dfrac{r^2}{R^2}.
  2. B=μ0Ir2πR2∝rB = \dfrac{\mu_0 I r}{2\pi R^2} \propto r.
  3. So the answer is (b) B∝rB \propto r.
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