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Question #43361

Difficulty: 8/10 | 🟢 Free

The Question

A solid rectangular conducting slab of width $W$ (along the x-axis), length $L$ (along the y-axis), and thickness $D$ (along the z-axis) moves with a constant velocity $\vec{v} = v \hat{i}$ . It passes through a region of non-uniform magnetic field $\vec{B} = \beta x^2 \hat{k}$ , where $\beta$ is a positive constant. Assume $L \gg W$ such that the induced eddy currents flow primarily along the y-axis, and edge effects at $y = \pm L/2$ can be neglected. At the instant the geometric center of the slab is at the origin $(0,0,0)$ , the total power dissipated in the solid slab due to eddy currents is $P_0$ . The solid slab is then replaced by a laminated slab of the exact same overall dimensions, consisting of $N=3$ identical insulated conducting strips, each of width $W/3$ (along the x-axis). What is the total power dissipated in the laminated slab at the same instant (when its geometric center is at the origin)?

Options

A.
$\frac{1}{9} P_0$
B.
$\frac{41}{81} P_0$
C.
$\frac{5}{81} P_0$
D.
$\frac{1}{3} P_0$

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

Power Dissipation Ratio|Eddy Currents Dissipation|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental origin of eddy currents and targets a common misconception regarding laminated cores.

Students often blindly apply the standard rule that laminating a core into $N$ segments reduces power dissipation by a factor of $1/N^2$ .

However, this rule is only valid when the induced electric field (or $dB/dt$ ) is uniform across the conductor.

In a non-uniform magnetic field gradient (here, $B \propto x^2$ ), the motional EMF is non-linear.

The electrostatic boundary conditions dictate that the local electrostatic field must cancel the average motional EMF within each specific isolated lamination.

Because the variance of $x^2$ over the displaced outer intervals is much higher than a linear function, the residual current density $J = \sigma(E_m - E_{es})$ is significantly larger in the outer laminations than the $1/N^2$ rule predicts, breaking the standard scaling law.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Power Dissipation Ratio
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free