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Question #43176

Difficulty: 8/10 | 🟢 Free

The Question

A region of vacuum contains a uniform static electric field E⃗0=E0i^\vec{E}_0 = E_0 \hat{i} and a uniform static magnetic field B⃗0=B0k^\vec{B}_0 = B_0 \hat{k}. The magnitudes of these fields are such that their energy densities in the vacuum are equal. A large flat slab of a linear diamagnetic material with relative permeability μr=0.80\mu_r = 0.80 and relative permittivity ϵr=1.0\epsilon_r = 1.0 is introduced into this region. The slab is oriented such that its two parallel faces are perpendicular to the zz-axis. Let uEu_E and uBu_B be the electric and magnetic energy densities inside the diamagnetic slab, respectively. What is the ratio uBuE\frac{u_B}{u_E}?

Options

  • A.

    54\frac{5}{4}

  • B.

    45\frac{4}{5}

  • C.

    11

  • D.

    2516\frac{25}{16}

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

Energy Density Partitioning|Magnitude Ratio Identity|Diamagnetic Property of Materials

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental definitions of electromagnetic energy density in linear media and the rigorous application of boundary conditions.

Students often rely on the heuristic that electric and magnetic energy densities are equal (uE=uBu_E = u_B), which holds for freely propagating EM waves in vacuum but not necessarily for static fields in a medium.

A lethal trap is incorrectly assuming the magnetic energy density formula is uB=12μB2u_B = \frac{1}{2}\mu B^2 (analogous to uE=12ϵE2u_E = \frac{1}{2}\epsilon E^2), which places μr\mu_r in the numerator and yields a ratio of 4/54/5.

Another trap is misapplying boundary conditions by assuming H⃗\vec{H} is continuous across a boundary perpendicular to B⃗\vec{B}, which also incorrectly yields 4/54/5.

By forcing the use of a diamagnetic medium (μr<1\mu_r < 1) with static crossed fields, the student must recognize that B⊥B_\perp is continuous and correctly apply uB=B22μu_B = \frac{B^2}{2\mu}, leading to an increased magnetic energy density relative to the vacuum state.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Waves - Class 12
ConceptEnergy Density Partitioning
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree