Start Learning Free

Before You Begin

Question #42687

Difficulty: 8/10 | 🟢 Free

The Question

A long cylindrical region of radius RR contains a time-varying magnetic field directed parallel to its central axis. The magnitude of the magnetic field is given by B(r,t)=βrtB(r,t) = \beta r t, where β\beta is a positive constant, rr is the radial distance from the axis, and tt is time. The magnetic field outside this region is zero. Calculate the total electrical energy stored in the induced electric field per unit length of the cylinder, within the region 0≤r≤R0 \le r \le R. (Assume ϵ0\epsilon_0 is the permittivity of free space).

Options

  • A.

    πϵ0β2R654\frac{\pi \epsilon_0 \beta^2 R^6}{54}

  • B.

    πϵ0β2R624\frac{\pi \epsilon_0 \beta^2 R^6}{24}

  • C.

    πϵ0β2R618\frac{\pi \epsilon_0 \beta^2 R^6}{18}

  • D.

    πϵ0β2R627\frac{\pi \epsilon_0 \beta^2 R^6}{27}

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 Locality|Induced Electric Fields|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of Faraday's Law with localized field energy density in a non-conservative setup.

The primary trap (Option B) catches students who incorrectly calculate magnetic flux by simply multiplying the local magnetic field B(r)B(r) by the area πr2\pi r^2, ignoring the spatial variation of the field inside the integral.

Another conceptual trap is attempting to define a scalar potential or use W=qΔVW = q\Delta V shortcuts, which fail because the induced electric field is non-conservative (∇×E⃗≠0\nabla \times \vec{E} \neq 0).

Students must rely on the fundamental energy density volume integral U=∫12ϵ0E2dτU = \int \frac{1}{2}\epsilon_0 E^2 d\tau.

Option C traps students who assume the electric field is uniform across the cross-section, and Option D catches those who forget the 1/21/2 factor in the energy density formula.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptEnergy Density Locality
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree