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JEE MainPhysicsMagnetic Effects of Current

A long straight solid cylindrical wire of radius R carries a steady current which is uniformly distributed across its cross-section. It is found that the magnitude of the magnetic field at a distance r₁ ( r₁ R ) from the axis. Which of the following relations is correct?

Options

  1. Ar₁ + r₂ = 2R
  2. Br₁ = r₂ R^2
  3. Cr₁^2 = r₂ R
  4. Dr₁ r₂ = R^2

Correct answer

D. r₁ r₂ = R^2

Step-by-step solution

Let the total steady current be I . For a point inside the solid cylinder ( r₁ B₁ = ₀ I r₁ 2 R^2 For a point outside the cylinder ( r₂ > R ), the magnetic field is: B₂ = ₀ I 2 r₂ Given that the magnitudes of the magnetic fields are equal ( B₁ = B₂ ): ₀ I r₁ 2 R^2 = ₀ I 2 r₂ Canceling the common terms ( ₀ , I , and 2 ) on both sides yields: r₁ R^2 = 1 r₂ Rearranging the equation, we get: r₁ r₂ = R^2 Answer: r₁ r₂ = R^2

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