To find the magnetic field $B$ at a radius $r = 2a$ from the axis of a long cylindrical conductor with a uniform current density, we can apply Ampère's law.

Step-by-Step Solution:

1. Given Information:

   • Radius of the conductor, $a = 0.55 \, \text{m}$
   • Current density, $J = 9.5 \, \text{A/m}^2$
   • Radius of the imaginary cylinder, $r = 2a = 2 \times 0.55 = 1.1 \, \text{m}$

2. Calculate the Total Current $I$: Since the current density $J$ is uniform, the total current $I$ flowing through the conductor can be calculated as: $I = J \cdot A$ where $A$ is the cross-sectional area of the conductor: $A = \pi a^2$ Thus, $I = J \cdot \pi a^2 = 9.5 \times \pi \times (0.55)^2$

3. Substitute and Calculate $I$: $I = 9.5 \times \pi \times 0.3025 \approx 9.019 \, \text{A}$

4. Apply Ampère's Law to Find $B$: Ampère's Law states: $\oint B \, \text{d}l = \mu_0 I_{\text{enc}}$ where $I_{\text{enc}}$ is the current enclosed within the path of radius $r$, and $\mu_0$ is the permeability of free space ($\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$).

   For $r = 2a$, which is outside the conductor, the entire current $I$ is enclosed. Since the magnetic field $B$ is constant over the circular path at radius $r$, we can simplify this to: $B \cdot (2\pi r) = \mu_0 I$ Solving for $B$: $B = \frac{\mu_0 I}{2\pi r}$

5. Substitute and Calculate $B$: $B = \frac{4\pi \times 10^{-7} \times 9.019}{2\pi \times 1.1}$ Simplify by canceling $\pi$ and calculating: $B \approx \frac{4 \times 10^{-7} \times 9.019}{2 \times 1.1} \approx 1.637 \times 10^{-6} \, \text{T}$

Final Answer:

$B \approx 1.64 \times 10^{-6} \, \text{T}$

Would you like more details, or do you have any further questions?

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Related Questions:

1. How does the magnetic field vary with distance inside the conductor?
2. What happens to the magnetic field at points beyond $r = 2a$?
3. How would the result change if the current density was not uniform?
4. How does Ampère's law apply to conductors with varying cross-sections?
5. How would the calculation change if the conductor were hollow?

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Tip:

For problems involving current distributions, always check if the current is uniformly distributed, as it simplifies calculations significantly with Ampère's law.