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Question #42813

Difficulty: 8/10 | 🟢 Free

The Question

A straight conducting rod of length LL has the shape of a truncated cone with circular end faces of radii aa and bb (b>ab > a). The rod is made of an isotropic material of uniform electrical resistivity ρ\rho. A constant potential difference VV is applied across its end faces. Assuming that the current density vector is uniform across any cross-section and parallel to the axis of the rod, which of the following expressions gives the magnitude of the electric field at the exact midpoint of the rod's axis?

Options

  • A.

    4VabL(a+b)2\frac{4 V a b}{L (a+b)^2}

  • B.

    VL\frac{V}{L}

  • C.

    V(a2+b2)2Lab\frac{V (a^2 + b^2)}{2 L a b}

  • D.

    2V(a2+b2)L(a+b)2\frac{2 V (a^2 + b^2)}{L (a+b)^2}

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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

Drift Velocity Mechanism|Potential Gradient Direction|Steady State Temperature Gradient

Why This Question Is Tricky

Common Mistakes

This question targets the common misconception that the electric field inside a current-carrying conductor is always uniform (E=V/LE = V/L).

Students often misapply this to non-uniform geometries, ignoring the continuity of steady current (I=J⋅A=constantI = J \cdot A = \text{constant}).

Because J=E/ρJ = E/\rho, the electric field must vary inversely with the cross-sectional area (E∝1/AE \propto 1/A).

By forcing the derivation of the spatial variation of E(x)E(x), the problem exposes the flaw of assuming linear potential drops or using average areas, highlighting that non-uniform conductors require surface charge distributions to maintain the steady-state current field.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptDrift Velocity Mechanism
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree