A long straight wire of a circular cross-section with radius ' a ' carries a steady current I . The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by

Magnetics PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q59 with Solution

View written solutionFree

Correct answer: B

For a long straight wire carrying steady current, the magnetic field at distance $r$ from the center is found using

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

Because of cylindrical symmetry, $B$ is constant on a circular Amperian loop of radius $r$ , so

$B(2\pi r)=\mu_0 I_{\text{enc}}$

Hence,

$B=\frac{\mu_0 I_{\text{enc}}}{2\pi r}$

Since current is uniformly distributed,

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

Current enclosed within radius $r$ is

$I_{\text{enc}}=J(\pi r^2)=\frac{I}{\pi a^2}\cdot \pi r^2=I\frac{r^2}{a^2}$

So,

=\frac{\mu_0 I}{2\pi a^2}r$$ Thus, for $r<a$, $$B\propto r$$ So the graph is a straight line starting from zero at $r=0$ and increasing linearly up to $r=a$. --- 3. **For points outside the wire: $r\ge a$** Now the entire current is enclosed: $$I_{\text{enc}}=I$$ Therefore, $$B=\frac{\mu_0 I}{2\pi r}$$ Thus, for $r>a$, $$B\propto \frac{1}{r}$$ So outside the wire, the graph decreases hyperbolically. --- 4. **Check continuity at $r=a$** Inside formula at $r=a$: $$B=\frac{\mu_0 I}{2\pi a^2}(a)=\frac{\mu_0 I}{2\pi a}$$ Outside formula at $r=a$: $$B=\frac{\mu_0 I}{2\pi a}$$ So the graph is continuous at $r=a$. --- 5. **Final graph shape** - From $r=0$ to $r=a$: straight line increasing linearly. - For $r>a$: decreases as $1/r$. - Continuous at $r=a$. This corresponds to **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** Derived answer: **B** They match.