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Question #42780

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The Question

A long straight wire carrying a steady current II along the positive zz-axis is placed in a vacuum (permeability μ0\mu_0). A uniform external magnetic field B⃗0=B0j^\vec{B}_0 = B_0 \hat{j} is also present throughout the space. Consider a cylindrical region of volume VV defined by the coordinates a≤r≤Ra \le r \le R, 0≤θ≤π/20 \le \theta \le \pi/2, and 0≤z≤L0 \le z \le L, where (r,θ,z)(r, \theta, z) are standard cylindrical coordinates with x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta. What is the total magnetic energy stored within this volume VV?

Options

  • A.

    μ0I2L16πln⁡(Ra)+πB02L8μ0(R2−a2)+IB0L2π(R−a)\frac{\mu_0 I^2 L}{16\pi} \ln\left(\frac{R}{a}\right) + \frac{\pi B_0^2 L}{8\mu_0} (R^2 - a^2) + \frac{I B_0 L}{2\pi} (R - a)

  • B.

    μ0I2L16πln⁡(Ra)+πB02L8μ0(R2−a2)\frac{\mu_0 I^2 L}{16\pi} \ln\left(\frac{R}{a}\right) + \frac{\pi B_0^2 L}{8\mu_0} (R^2 - a^2)

  • C.

    μ0I2L16πln⁡(Ra)+πB02L8μ0(R2−a2)−IB0L2π(R−a)\frac{\mu_0 I^2 L}{16\pi} \ln\left(\frac{R}{a}\right) + \frac{\pi B_0^2 L}{8\mu_0} (R^2 - a^2) - \frac{I B_0 L}{2\pi} (R - a)

  • D.

    μ0I2L8πln⁡(Ra)+πB02L4μ0(R2−a2)+IB0Lπ(R−a)\frac{\mu_0 I^2 L}{8\pi} \ln\left(\frac{R}{a}\right) + \frac{\pi B_0^2 L}{4\mu_0} (R^2 - a^2) + \frac{I B_0 L}{\pi} (R - a)

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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|Gauss Law Applicability|Superposition of Magnetic Fields

Why This Question Is Tricky

Common Mistakes

Students incorrectly assume that the total magnetic energy of a system is simply the sum of the individual self-energies of the constituent fields (U=U1+U2U = U_1 + U_2).

They fail to recognize that energy density is proportional to the square of the net field (u∝∣B⃗net∣2u \propto |\vec{B}_{net}|^2), which introduces a non-linear cross-term B⃗1⋅B⃗2\vec{B}_1 \cdot \vec{B}_2.

While this interaction energy often integrates to zero over all space due to symmetry, restricting the integration volume to a specific quadrant prevents the angular integral from vanishing.

Ignoring this 'locality' constraint of energy density leads to missing the interaction term entirely.

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