A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distr

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A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across the cross-section. The magnetic field is:

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Given: A long straight wire of radius a carries a steady current I uniformly distributed over its cross-section. Find: How the magnetic field varies with distance r from the axis in the regions r<a and ra. Using Ampere's law, the magnetic field at distance r depends on the current enclosed by an Amperian circle of radius r. For the region inside the wire where r<a, the current density is uniform, so the enclosed current is proportional to the enclosed area: I enc = I r^2 a^2 = I r^2a^2 Applying Ampere's law, B(2 r) = 0 I enc = 0 I r^2a^2 Hence, B = 0 I2 a^2r So, inside the wire, B r. For the region outside the wire where ra, the entire current I is enclosed. Therefore, B(2 r) = 0 I which gives B = 0 I2 r So, outside the wire, B 1r. Therefore, the magnetic field is directly proportional to r in the region r<a and inversely proportional to r in the region ra. The correct option is C.

Given: The current is uniformly distributed over the circular cross-section of radius $a$ .

Find: The radial dependence of magnetic field $B$ .

The uniform current density is

J = \frac{I}{\pi a^2}

At a distance ** $r**$

Now by Ampere's law,

\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}

By symmetry, $B$ is constant on a circular path of radius $r$ , so

B(2\pi r)=\mu_0 I\frac{r^2}{a^2}

Thus,

B=\frac{\mu_0 I}{2\pi a^2}r

For $r>a$ , the enclosed current becomes the full current $I$ . Hence,

B(2\pi r)=\mu_0 I

which gives

B=\frac{\mu_0 I}{2\pi r}

Thus the variation is $B\propto r$ for $r<a$ and $B\propto \frac{1}{r}$ for $r>a$ . So the correct option is C.

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