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Question #43030

Difficulty: 8/10 | 🟢 Free

The Question

A small rigid circular loop of radius $R$ and mass $M$ carries a steady current $I$ . It is placed with its center at the origin $(0,0,0)$ of a Cartesian coordinate system, lying in the $y$ - $z$ plane such that its magnetic dipole moment $\vec{\mu}$ is directed along the $+x$ -axis. The region contains a static, non-uniform magnetic field given by $\vec{B} = B_0 (\alpha z \hat{i} + \alpha x \hat{k} + \hat{k})$ , where $B_0$ and $\alpha$ are positive constants. If the loop is released from rest, what are its initial translational acceleration $\vec{a}$ and initial angular acceleration $\vec{\alpha}_{ang}$ ? (Assume gravity is negligible).

Options

A.
$\vec{a} = \frac{\pi I R^2 B_0 \alpha}{M} \hat{k}, \quad \vec{\alpha}_{ang} = -\frac{2 \pi I B_0}{M} \hat{j}$
B.
$\vec{a} = \vec{0}, \quad \vec{\alpha}_{ang} = -\frac{2 \pi I B_0}{M} \hat{j}$
C.
$\vec{a} = \frac{\pi I R^2 B_0 \alpha}{M} \hat{i}, \quad \vec{\alpha}_{ang} = \frac{2 \pi I B_0}{M} \hat{j}$
D.
$\vec{a} = \frac{\pi I R^2 B_0 \alpha}{M} \hat{k}, \quad \vec{\alpha}_{ang} = -\frac{\pi I B_0}{M} \hat{j}$

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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

Torque on Current Loop|Potential vs Field Relationship|Dipole in Non-Uniform Field|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the behavior of a magnetic dipole in a non-uniform magnetic field.

Students are heavily conditioned by introductory problems to assume that a magnetic dipole only experiences a torque ( $\vec{\tau} = \vec{\mu} \times \vec{B}$ ) and that the net translational force is zero.

However, this is only true in a strictly uniform magnetic field.

By introducing a spatial gradient ( $\alpha$ ), the field becomes non-uniform, inducing a net translational force $\vec{F} = \nabla(\vec{\mu} \cdot \vec{B})$ .

Students who fail to recognize this will incorrectly select Option B, assuming zero translational acceleration.

Furthermore, Option D traps students who incorrectly use the moment of inertia of a ring about its central axis ( $MR^2$ ) instead of its diameter ( $\frac{1}{2}MR^2$ ), which is the actual axis of rotation induced by the torque.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Torque on Current Loop
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free