Question

A long coaxial cable carries current 'I' (current flows down the surface of inner cylinder of radius 'r1' and back along the outer cylinder of radius 'r2'). The magnetic energy stored in a section of length 'L' is

• A. \( \frac{\mu_0}{4\pi} I^2 L \ln\left(\frac{r_1}{r_2}\right) \)
• B. \( \frac{\mu_0}{4\pi} I L \ln\left(\frac{r_2}{r_1}\right) \)
• C. \( \frac{\mu_0}{4\pi} I^2 L \ln\left(\frac{r_2}{r_1}\right) \)
• D. \( \frac{\epsilon_0}{4\pi} I^2 L \ln\left(\frac{r_2}{r_1}\right) \)

Hint:

For problems involving energy in fields (electric or magnetic), the standard procedure is: 1. Find the field (\(\vec{E}\) or \(\vec{B}\)) using Gauss's Law or Ampere's Law. 2. Calculate the energy density (\(u_E = \frac{1}{2}\epsilon_0 E^2\) or \(u_m = \frac{1}{2\mu_0} B^2\)). 3. Integrate the energy density over the relevant volume.

Answer 1

The Correct Option is C

Step 1: Conceptual Foundation: The magnetic energy stored in a system is directly proportional to the magnetic field it generates. This problem requires determining the magnetic field within the region between coaxial cylinders and subsequently integrating the magnetic energy density over this volume.
Step 2: Core Formula/Methodology: The magnetic energy \( U_m \) within a volume \( V \) is computed by integrating the magnetic energy density \( u_m \):
\[ U_m = \int_V u_m dV \]The magnetic energy density is defined as:
\[ u_m = \frac{B^2}{2\mu_0} \]Ampere's law will be employed to ascertain the magnetic field \( B \).
Step 3: Detailed Derivation:
1. Magnetic Field Calculation: For a circular Amperian loop of radius \( r \) where \( r_1<r<r_2 \), the enclosed current is 'I'. Applying Ampere's Law:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]This yields:
\[ B(2\pi r) = \mu_0 I \]And thus:
\[ B = \frac{\mu_0 I}{2\pi r} \]The magnetic field is absent for \( r<r_1 \) and \( r>r_2 \).
2. Magnetic Energy Density Computation: Substituting the derived magnetic field:
\[ u_m = \frac{B^2}{2\mu_0} = \frac{1}{2\mu_0} \left( \frac{\mu_0 I}{2\pi r} \right)^2 = \frac{\mu_0 I^2}{8\pi^2 r^2} \]3. Total Magnetic Energy Integration: The energy density is integrated over the volume between the cylinders for a length 'L'. The volume element \( dV \) in cylindrical coordinates is \( (2\pi r) L dr \). The total magnetic energy is:
\[ U_m = \int_{r_1}^{r_2} u_m dV = \int_{r_1}^{r_2} \left( \frac{\mu_0 I^2}{8\pi^2 r^2} \right) (2\pi r L dr) \]Simplifying the integral:
\[ U_m = \frac{\mu_0 I^2 L}{4\pi} \int_{r_1}^{r_2} \frac{1}{r} dr \]Evaluating the integral:
\[ U_m = \frac{\mu_0 I^2 L}{4\pi} [\ln(r)]_{r_1}^{r_2} \]This results in:
\[ U_m = \frac{\mu_0 I^2 L}{4\pi} (\ln(r_2) - \ln(r_1)) \]Step 4: Final Result: Utilizing the logarithmic property \( \ln(a) - \ln(b) = \ln(a/b) \):
\[ U_m = \frac{\mu_0 I^2 L}{4\pi} \ln\left(\frac{r_2}{r_1}\right) \]This result corresponds to option (C).