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JEE Advanced 2023 Paper 2 · Physics · Magnetic Effects of Current · hard

A long solid cylindrical conductor of radius a=10 cm carries a uniform current density J=5×10^5 A m^-2 along its axis. It has a long cylindrical hole…

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A long solid cylindrical conductor of radius a=10 cma=10\ \mathrm{cm} carries a uniform current density J=5×105 A m−2J=5\times10^{5}\ \mathrm{A\,m^{-2}} along its axis. It has a long cylindrical hole of radius b=1 cmb=1\ \mathrm{cm} whose axis is parallel to the axis of the conductor and at a distance d=4 cmd=4\ \mathrm{cm} from it; the current density in the remaining material is unchanged. Take μ0=4π×10−7 T m A−1\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}. Which of the following statements is/are correct?
  1. A. The magnetic field inside the hole is uniform, of magnitude 12.6 mT12.6\ \mathrm{mT}.
  2. B. The magnetic field at the point on the outer surface of the conductor that is farthest from the hole is 31.2 mT31.2\ \mathrm{mT}.
  3. C. The magnetic field on the axis of the conductor is zero.
  4. D. The magnetic field inside the hole is parallel to the line joining the two axes.
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Answer: A, B

Use superposition: the actual conductor = full cylinder with current density +J+J plus a cylinder filling the hole with −J-J. For a cylinder, inside B⃗=12μ0J⃗×r⃗\vec B=\tfrac12\mu_0\vec J\times\vec r (from its axis), outside B=μ0Jρ22rB=\dfrac{\mu_0J\rho^2}{2r}. (A) Inside the hole B⃗=12μ0J⃗×(r⃗1−r⃗2)=12μ0J⃗×d⃗\vec B=\tfrac12\mu_0\vec J\times(\vec r_1-\vec r_2)=\tfrac12\mu_0\vec J\times\vec d, uniform, magnitude 12μ0Jd=12(4π×10−7)(5×105)(0.04)=12.6 mT\tfrac12\mu_0Jd=\tfrac12(4\pi\times10^{-7})(5\times10^5)(0.04)=12.6\ \mathrm{mT}: correct. (B) At the farthest surface point, Bfull=12μ0Ja=31.4 mTB_{full}=\tfrac12\mu_0Ja=31.4\ \mathrm{mT} and the hole contributes −μ0Jb22(a+d)=−0.22 mT-\dfrac{\mu_0Jb^2}{2(a+d)}=-0.22\ \mathrm{mT} (distance a+d=14 cma+d=14\ \mathrm{cm}), giving 31.2 mT31.2\ \mathrm{mT}: correct. (C) On the axis the full cylinder gives zero but the hole gives μ0Jb22d=0.79 mT\dfrac{\mu_0Jb^2}{2d}=0.79\ \mathrm{mT}: incorrect. (D) J⃗×d⃗\vec J\times\vec d is perpendicular to d⃗\vec d, so the field in the hole is perpendicular to the line of axes: incorrect.
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