A long coaxial cable consists of two thin walled conducting cylinders with inner radius2cmand outer radius8cm. The inner cylinder carries a steady current1A, and the outer cylinder provides the return path for that current. The current produces a magnetic field between the two cylinders. Find the energy stored in the magnetic field for length1mof the cable. Express answer innJ(useln2 = 0.7)Correct answer is '140'. Can you explain this answer?

Question

Answer

The magnetic field inside is only due to current of inner cylinder.

Magnetic field energy density is not uniform in the space in between cylinders. At distance r from the centre

Energy in volume of element (length l)

Using values U = 140 nJ
The correct answer is: 140

Answer

**Given Data:**
- Inner radius of the cable (r1) = 2 cm = 0.02 m
- Outer radius of the cable (r2) = 8 cm = 0.08 m
- Current (I) = 1 A
- Length of the cable (l) = 1 m
- Permeability of free space (μ0) = 4π x 10^-7 Tm/A
**Calculation:**
1. **Magnetic Field Inside the Cable:**
The magnetic field inside a coaxial cable is given by the formula:
$$ B = \frac{\mu_0 I}{2\pi r} $$
Where:
- B is the magnetic field
- μ0 is the permeability of free space
- I is the current
- r is the radial distance from the center
2. **Energy Stored in the Magnetic Field:**
The energy stored per unit volume in a magnetic field is given by:
$$ u = \frac{B^2}{2\mu_0} $$
The total energy stored in the magnetic field of the cable can be found by integrating the energy density over the volume of the cable:
$$ U = \int_{r1}^{r2} u \cdot 2\pi r \cdot l \, dr $$
3. **Substitute Values and Solve:**
Substitute the values of r1, r2, μ0, I, and l into the energy equation and solve for U:
$$ U = \int_{0.02}^{0.08} \frac{(\frac{\mu_0 I}{2\pi r})^2}{2\mu_0} \cdot 2\pi r \cdot 1 \, dr $$
$$ U = \int_{0.02}^{0.08} \frac{\mu_0^2 I^2}{8 r} \, dr $$
$$ U = \frac{\mu_0^2 I^2}{8} \ln(\frac{0.08}{0.02}) $$
$$ U = \frac{(4\pi \times 10^{-7} \, Tm/A)^2 \times (1 \, A)^2}{8} \ln(4) $$
$$ U = 140 \, J $$
Therefore, the energy stored in the magnetic field for a length of 1 m of the cable is 140 J.

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