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Question #42725

Difficulty: 8/10 | 🟢 Free

The Question

A uniform heavy wire of un-stretched length $L$ , mass $M$ , initial cross-sectional area $A_0$ , and resistivity $\rho$ hangs vertically. It is initially supported along its entire length such that its strain is zero everywhere. The support is suddenly removed, allowing the wire to fall and stretch under its own weight. A constant direct current $I$ flows downwards through the wire. The wire is also immersed in an external radial magnetic field $\vec{B} = B_0 \hat{r}$ , which exerts an azimuthal torque on the wire; assume the resulting torsional deformation is negligible and does not affect the wire's dimensions. Assuming the volume of each infinitesimal element of the wire remains constant during elastic deformation, and the material obeys Hooke's law with Young's modulus $Y$ , what is the exact instantaneous electrical resistance of the wire at the instant of its maximum downward displacement?

Options

A.
$\frac{\rho L}{A_0} \left( 1 + \frac{2 M g}{A_0 Y} + \frac{4 M^2 g^2}{3 A_0^2 Y^2} \right)$
B.
$\frac{\rho L}{A_0} \left( 1 + \frac{M g}{A_0 Y} + \frac{M^2 g^2}{3 A_0^2 Y^2} \right)$
C.
$\frac{\rho L}{A_0} \left( 1 + \frac{2 M g}{A_0 Y} \right)$
D.
$\frac{\rho L}{A_0} \left( 1 + \frac{M g}{A_0 Y} \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Current Density Vector|Wire Elongation under Self-Weight|Work-Energy Theorem for Rotation|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of continuum mechanics and electrodynamics.

A primary trap is assuming the maximum dynamic elongation equals the static equilibrium elongation, ignoring the Work-Energy Theorem which dictates that a suddenly released elastic body stretches twice as much as its static equilibrium position (since $W_{gravity} = \Delta U_{elastic}$ at maximum stretch).

Furthermore, students often miscalculate the new resistance by only accounting for the length increase ( $R \propto L$ ) and ignoring the local cross-sectional area reduction (Poisson effect), or by using an average strain instead of integrating the non-uniform current density $J(y)$ and local resistance $dR = \rho dy' / A(y)$ over the stress gradient.

The radial magnetic field serves as a conceptual distractor to test confidence in identifying relevant forces.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Current Density Vector
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free