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Question #43393

Difficulty: 8/10 | 🟢 Free

The Question

Consider two independent physical systems in a vacuum:

System I: A thick spherical shell of inner radius RR and outer radius 2R2R has a non-uniform mass density ρ(r)=ρ0rR\rho(r) = \rho_0 \frac{r}{R}, where rr is the radial distance from the center and ρ0\rho_0 is a constant in kg m−3\text{kg m}^{-3}.

System II: A very long thick-walled cylindrical solenoid of inner radius RR and outer radius 2R2R carries a non-uniform azimuthal volume current density J(r)=J0rRJ(r) = J_0 \frac{r}{R}, where rr is the radial distance from the central axis and J0J_0 is a constant in A m−2\text{A m}^{-2}.

Let EgE_g be the magnitude of the gravitational field at a distance r=R/2r = R/2 from the center of the spherical shell. Let BB be the magnitude of the magnetic field at a distance r=R/2r = R/2 from the central axis of the solenoid. Which of the following correctly gives EgE_g and BB?

Options

  • A.

    Eg=0E_g = 0, B=32μ0J0RB = \frac{3}{2} \mu_0 J_0 R

  • B.

    Eg=0E_g = 0, B=0B = 0

  • C.

    Eg=14πGρ0RE_g = \frac{1}{4} \pi G \rho_0 R, B=32μ0J0RB = \frac{3}{2} \mu_0 J_0 R

  • D.

    Eg=14πGρ0RE_g = \frac{1}{4} \pi G \rho_0 R, B=0B = 0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Ampere’s Circuital Law Symmetry|Gravitational Field Variation (Depth vs Height)|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the misapplication of the 'shell theorem' from gravitation to magnetostatics.

Students often assume that inside any hollow cavity of a symmetric distribution, the field is zero.

While true for spherical mass shells due to the inverse-square law and cancellation geometry (Gauss's Law), it fails for a thick-walled solenoid.

In magnetostatics, the circulation geometry (Ampere's Law) dictates that the axial field inside the cavity depends on the total azimuthal current integrated across the entire thickness of the shell, resulting in a non-zero, uniform internal field.

The trap exploits the heuristic 'empty cavity equals zero field'.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptAmpere’s Circuital Law Symmetry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree