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Question #42975

Difficulty: 8/10 | 🟢 Free

The Question

A rigid conducting rod of length $L$ and uniform electrical resistance $R$ rotates in a horizontal plane with a constant angular velocity $\omega$ about a vertical axis passing through one of its ends ( $r=0$ ). The other end of the rod slides without friction on a stationary, perfectly conducting circular ring of radius $L$ . A stationary, perfectly conducting wire connects the pivot to the ring, completing the electrical circuit. A non-uniform, time-independent magnetic field is applied perpendicular to the plane of rotation. The magnitude of the magnetic field varies with the radial distance $r$ from the pivot as $B(r) = B_0 \left(1 - \frac{r}{L}\right)$ , where $B_0$ is a positive constant. Neglecting the self-inductance of the loop, what is the magnitude of the external mechanical torque $\tau_{\text{ext}}$ that must be applied to the rod to maintain its constant angular velocity $\omega$ ?

Options

A.
$\frac{B_0^2 \omega L^4}{36 R}$
B.
$\frac{B_0^2 \omega L^4}{16 R}$
C.
$\frac{B_0^2 \omega L^4}{24 R}$
D.
$\frac{B_0^2 \omega L^4}{4 R}$

Check Your Answer

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

Faraday’s Law Vector Form|Motional EMF in Rotating Rods|Moving Charge in Non-Uniform B|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate motional EMF and magnetic torque in a non-uniform magnetic field, exposing two critical traps.

First, students frequently substitute the spatial average of the magnetic field ( $B_{\text{avg}} = B_0/2$ ) directly into the standard uniform-field formula $\varepsilon = \frac{1}{2} B \omega L^2$ , yielding an incorrect EMF and a torque of $\frac{B_0^2 \omega L^4}{16 R}$ (Option B).

This violates Faraday's Law because the local field must be weighted by the local velocity ( $\omega r$ ) during integration.

Second, even if students correctly integrate to find the induced current, they often incorrectly calculate the magnetic torque by assuming the net magnetic force acts at the rod's center of mass ( $r=L/2$ ) with an average field, leading to $\frac{B_0^2 \omega L^4}{24 R}$ (Option C).

The problem enforces the rigorous application of calculus for both electrical and mechanical domains, or the elegant use of energy conservation ( $P_{\text{ext}} = I^2 R$ ) to bypass the torque integration entirely.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	Faraday’s Law Vector Form
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free