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Question #43024

Difficulty: 8/10 | 🟢 Free

The Question

Two infinitely long parallel wires, each of radius $a$ , carry constant currents $I$ in opposite directions. The distance between their central axes is $x$ (where $x \gg a$ ). The magnetic force per unit length $F$ between the wires can be derived using the principle of virtual work by considering the variation of the magnetic interaction energy per unit length $U_{int}(x)$ with respect to $x$ . The interaction energy per unit length is defined as $U_{int} = \int \frac{\vec{B}_1 \cdot \vec{B}_2}{\mu_0} dA$ , where the integral is over the entire 2D plane perpendicular to the wires. Which of the following expressions correctly represents the $x$ -dependent part of $U_{int}(x)$ and the correct thermodynamic relation used to find the force $F$ (where $F > 0$ denotes a repulsive force)? Let $C$ be a constant independent of $x$ .

Options

A.
$U_{int}(x) = \frac{\mu_0 I^2}{2\pi} \ln(x) + C$ , and $F = +\frac{\partial U_{int}}{\partial x}$
B.
$U_{int}(x) = -\frac{\mu_0 I^2}{2\pi} \ln(x) + C$ , and $F = -\frac{\partial U_{int}}{\partial x}$
C.
$U_{int}(x) = \frac{\mu_0 I^2}{4\pi} \ln(x) + C$ , and $F = +\frac{\partial U_{int}}{\partial x}$
D.
$U_{int}(x) = \frac{\mu_0 I^2}{2\pi} \ln(x) + C$ , and $F = -\frac{\partial U_{int}}{\partial x}$

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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

Force Between Parallel Conductors|Energy Density Locality|Integration

Why This Question Is Tricky

Common Mistakes

This question tests two advanced concepts in electromagnetism: the locality of magnetic energy density and the principle of virtual work for constant-current systems.

First, students often incorrectly assume the interaction energy density is $\frac{\vec{B}_1 \cdot \vec{B}_2}{2\mu_0}$ by blindly applying the $1/2$ factor from the self-energy terms ( $B^2/2\mu_0$ ), which leads to a factor of $1/2$ error in $U_{int}$ .

The correct cross-term from expanding $(\vec{B}_1+\vec{B}_2)^2/2\mu_0$ is exactly $\frac{\vec{B}_1 \cdot \vec{B}_2}{\mu_0}$ .

Second, students habitually apply the conservative force relation $F = -\nabla U$ from mechanics.

However, for a magnetic system maintained at constant current, the power supply does work equal to twice the change in magnetic energy ( $dW_{batt} = 2dU$ ).

Consequently, the mechanical work done is $dW_{mech} = +dU$ , leading to $F = +\nabla U$ .

Applying the incorrect negative sign yields an attractive force, prompting some students to artificially flip the sign of the interaction energy to force a repulsive result.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Force Between Parallel Conductors
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free