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: Understanding the Concept:
The magnetic energy stored in a system is related to the magnetic field it produces. We need to find the magnetic field in the region between the coaxial cylinders and then integrate the magnetic energy density over the volume of this region.
Step 2: Key Formula or Approach:
The magnetic energy \( U_m \) stored in a volume \( V \) is given by the integral of the magnetic energy density \( u_m \):
\[ U_m = \int_V u_m dV \] where the energy density is:
\[ u_m = \frac{B^2}{2\mu_0} \] We will use Ampere's law to find the magnetic field \( B \).
Step 3: Detailed Explanation:
1. Find the magnetic field (B):
Consider a circular Amperian loop of radius \( r \) such that \( r_1<r<r_2 \). The current enclosed by this loop is 'I'. According to Ampere's Law:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \] \[ B(2\pi r) = \mu_0 I \] \[ B = \frac{\mu_0 I}{2\pi r} \] The magnetic field is zero for \( r<r_1 \) and \( r>r_2 \).
2. Calculate the magnetic energy density (\( u_m \)):
\[ u_m = \frac{B^2}{2\mu_0} = \frac{1}{2\mu_0} \left( \frac{\mu_0 I}{2\pi r} \right)^2 = \frac{\mu_0 I^2}{8\pi^2 r^2} \] 3. Integrate to find the total magnetic energy (\( U_m \)):
We integrate the energy density over the volume between the cylinders for a length 'L'. The volume element \( dV \) in cylindrical coordinates is \( (2\pi r) L dr \).
\[ U_m = \int_{r_1}^{r_2} u_m dV = \int_{r_1}^{r_2} \left( \frac{\mu_0 I^2}{8\pi^2 r^2} \right) (2\pi r L dr) \] \[ U_m = \frac{\mu_0 I^2 L}{4\pi} \int_{r_1}^{r_2} \frac{1}{r} dr \] \[ U_m = \frac{\mu_0 I^2 L}{4\pi} [\ln(r)]_{r_1}^{r_2} \] \[ U_m = \frac{\mu_0 I^2 L}{4\pi} (\ln(r_2) - \ln(r_1)) \] Step 4: Final Answer:
Using the property of logarithms, \( \ln(a) - \ln(b) = \ln(a/b) \), we get:
\[ U_m = \frac{\mu_0 I^2 L}{4\pi} \ln\left(\frac{r_2}{r_1}\right) \] This matches option (C).