Before You Begin

Question #42669

Difficulty: 8/10 | 🟢 Free

The Question

A rigid, non-conducting circular loop of radius $R$ is free to rotate in three dimensions about its center, which is fixed at the origin. A thin wire carrying a steady current $I$ is wound around the loop. Two small point charges, $+q$ and $-q$ , are fixed at diametrically opposite points on the loop's circumference. The system is placed in a region with a uniform magnetic field $\vec{B} = B_0 \hat{k}$ and an electric field described by the equipotential surfaces $V(x,y,z) = -E_0 z$ . What is the minimum work required by an external agent to rotate the loop slowly from its most stable equilibrium orientation to its most unstable equilibrium orientation?

Options

A.
$2 \sqrt{\pi^2 I^2 R^4 B_0^2 + 4 q^2 R^2 E_0^2}$
B.
$2 (\pi I R^2 B_0 + 2 q R E_0)$
C.
$2 \sqrt{\pi^2 I^2 R^4 B_0^2 + q^2 R^2 E_0^2}$
D.
$2 |\pi I R^2 B_0 - 2 q R E_0|$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Equipotential Surface Geometry|Torque on Current Loop|Magnetic Dipole Moment Orientation|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

The core trap in this question is the assumption that independent degrees of freedom exist for the alignment of the magnetic and electric dipoles.

Students often calculate the minimum potential energy by assuming both $\vec{m}$ and $\vec{p}$ can simultaneously align perfectly with their respective fields, yielding $U_{min} = -mB - pE$ (Option B).

This ignores the rigid body constraint that $\vec{m}$ (normal to the loop) and $\vec{p}$ (in the plane of the loop) are structurally orthogonal ( $\vec{m} \cdot \vec{p} = 0$ ).

When placed in parallel external fields, the dipoles are forced into a 'lethal collision' of constraints, competing for alignment.

The system must settle into a compromised equilibrium governed by $U(\theta) = -mB\cos\theta - pE\sin\theta$ , requiring vector addition of the energy coefficients rather than simple scalar addition.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Equipotential Surface Geometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free