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Step-by-Step Solution
Step 1: State the Problem Clearly
We have a long, thick current-carrying cable of radius $R$ that carries a total current $I$ uniformly distributed across its circular cross-section. We want to determine how the magnetic field $B(r)$ varies for different distances $r$ from the center of the cable.
Step 2: Apply Ampère’s Circuital Law
Ampère’s law states:
$$
\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}},
$$
where $I_{\text{enc}}$ is the current enclosed by the chosen Amperian loop.
Step 3: Find $B(r)$ for $r \leq R$ (Inside the Cable)
1. Consider a circular Amperian loop of radius $r$ (where $r \le R$) centered on the axis of the cable.
2. The total current $I$ is uniformly distributed across the cross-sectional area $\pi R^2$. Thus, the current density $J$ is:
$$
J = \frac{I}{\pi R^2}.
$$
3. The current enclosed by the Amperian loop of radius $r$ is then:
$$
I_{\text{enc}} = J \times (\text{area of cross-section up to } r) = \frac{I}{\pi R^2} \times \pi r^2 = I \left(\frac{r^2}{R^2}\right).
$$
4. By symmetry, $B(r)$ is constant on the circular path of radius $r$, so Ampère’s law becomes:
$$
B(r) \cdot 2\pi r = \mu_0 \cdot I_{\text{enc}} = \mu_0 \cdot I \left(\frac{r^2}{R^2}\right).
$$
5. Solving for $B(r)$ inside the cable:
$$
B_{in}(r) = \frac{\mu_0 I}{2\pi R^2} \, r.
$$
This shows that $B_{in}$ increases linearly with $r$ from the center ($r=0$) to $r=R$.
Step 4: Find $B(r)$ for $r \geq R$ (Outside the Cable)
1. For $r \ge R$, the Amperian loop encloses the entire current $I$.
2. By Ampère’s law:
$$
B(r) \cdot 2\pi r = \mu_0 I.
$$
3. Hence, the magnetic field outside the cable is:
$$
B_{out}(r) = \frac{\mu_0 I}{2\pi r}.
$$
This shows that $B_{out}$ decreases inversely with $r$ outside the cable.
Step 5: Combine the Two Regions and Interpret the Graph
• For $0 \le r \le R$, $B(r) \propto r$, so it rises linearly with $r$.
• For $r > R$, $B(r) \propto \frac{1}{r}$, so it decreases as $r$ increases.
The correct graphical representation that matches these results is:
Step 6: Final Expressions
$$
B_{in}(r) = \frac{\mu_0 I r}{2\pi R^2} \quad \text{for} \quad r \le R,
$$
$$
B_{out}(r) = \frac{\mu_0 I}{2\pi r} \quad \text{for} \quad r \ge R.
$$
Hence, the option with the piecewise linear (inside) and inverse (outside) dependence on $r$ is correct.
Reference Image from Given Solution
