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Step-by-Step Solution
Step 1: Understand the Physical Situation
A long cylindrical conductor of radius $R$ carries a steady current $I$. We wish to understand how the magnitude of the magnetic field $B$ varies with the distance $d$ from the center of the conductor, both inside ($d \le R$) and outside ($d > R$) the conductor.
Step 2: Magnetic Field Inside the Conductor ($d \le R$)
Inside the conductor, the current is assumed uniformly distributed across its cross-sectional area. By applying Ampère’s circuital law (in integral form) for a circular loop of radius $d$ (where $d \le R$) concentric with the axis of the cylinder, we get:
$ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \, I_{\text{enc}} $
The enclosed current $I_{\text{enc}}$ for radius $d$ is proportional to the area enclosed, so:
$ I_{\text{enc}} = I \times \frac{\pi d^2}{\pi R^2} = I \frac{d^2}{R^2}.
$
Since $B$ is constant along the chosen circular path and the path length is $2 \pi d$, Ampère’s law becomes:
$ B \cdot (2 \pi d) = \mu_0 \left(I \frac{d^2}{R^2}\right).
$
Solving for $B$ inside the conductor:
$ B = \frac{\mu_0 \, I}{2\pi R^2} \, d.
$
This shows that $B$ is directly proportional to $d$ inside the conductor ($B \propto d$). Hence, the plot of $B$ versus $d$ is a straight line passing through the origin for $d \le R$.
Step 3: Magnetic Field on the Surface ($d = R$)
At the surface ($d = R$), we substitute $d = R$ into the expression for $B_{\text{inside}}$:
$ B_{\text{surface}} = \frac{\mu_0 \, I}{2 \pi R^2} \cdot R = \frac{\mu_0 \, I}{2 \pi R}.
$
Step 4: Magnetic Field Outside the Conductor ($d > R$)
Outside the conductor, the entire current $I$ is enclosed by the circular path of radius $d$. Applying Ampère’s circuital law again for $d > R$ gives:
$ B \cdot (2 \pi d) = \mu_0 \, I.
$
Solving for $B$ outside the conductor:
$ B = \frac{\mu_0 \, I}{2 \pi d}.
$
This shows that for $d > R$, $B \propto \frac{1}{d}$. Therefore, the field decreases inversely with distance beyond the radius of the conductor.
Step 5: Final Graphical Representation
Combining these results:
Inside ($0 \le d \le R$): $B \propto d$ (a straight line through the origin).
Outside ($d > R$): $B \propto \frac{1}{d}$ (a hyperbolic decrease).
The correct plot (from the given options) shows $B$ increasing linearly from the center up to $d=R$, and then decreasing as $1/d$ for $d>R$. Thus, the correct figure is:
