Before You Begin

Question #42949

Difficulty: 8/10 | 🟢 Free

The Question

A conducting rod of length $L$ is pivoted at one end $O$ and rotates with a constant angular velocity $\omega$ in a horizontal plane. Its free end makes sliding contact with a stationary, perfectly conducting circular ring of radius $L$ centered at $O$ . A stationary straight conducting wire connects $O$ to a fixed point $P$ on the ring. The entire system is placed in a magnetic field perpendicular to the plane of the ring. The magnetic field is a superposition of two components:

A uniform, constant magnetic field of magnitude $B_0$ present everywhere.
A time-varying magnetic field of magnitude $B_1(t) = \alpha t$ (where $\alpha > 0$ ) present only within a cylindrical region of radius $R$ ( $R < L$ ) centered at $O$ . Both magnetic fields point in the same direction. At $t=0$ , the rotating rod coincides with the fixed wire $OP$ . What is the magnitude of the total induced electromotive force (EMF) in the closed loop formed by the rod, the fixed wire $OP$ , and the arc of the ring swept by the rod at time $t$ ?

Options

A.
$\frac{1}{2} \omega (B_0 L^2 + \alpha R^2 t)$
B.
$\frac{1}{2} \omega (B_0 L^2 + 2\alpha R^2 t)$
C.
$\frac{1}{2} \omega (B_0 L^2 + 2\alpha L^2 t)$
D.
$\frac{1}{2} \omega (B_0 L^2 - \alpha R^2 t)$

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

Superposition of Magnetic Fields|Lenz’s Law Energy Conservation|Motional EMF in Rotating Rods|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the Superposition Principle of Induced EMF and the common misapplication of the motional EMF formula.

Students often calculate the induced EMF by only integrating $\int (\vec{v} \times \vec{B}) \cdot d\vec{l}$ along the moving rod, yielding $\varepsilon_{mot} = \frac{1}{2}\omega(B_0 L^2 + \alpha R^2 t)$ .

They fall into the trap of failing to account for the transformer EMF ( $\oint \vec{E} \cdot d\vec{l}$ ) induced by the time-varying magnetic field component ( $\partial \vec{B}/\partial t$ ).

By correctly applying Faraday's Law ( $\varepsilon = -d\Phi/dt$ ) on the expanding sector, both the motional component (due to $dA/dt$ ) and the transformer component (due to $dB/dt$ ) are naturally incorporated via the product rule, revealing the missing factor of 2 in the time-varying term.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Magnetism and Matter - Class 12
Concept	Superposition of Magnetic Fields
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free