1. Given: A long straight wire of radius $R$ carries a steady current $I$, uniformly distributed over its cross-section. We need the magnetic field at a distance $r$ from the centre, where $r<R$.

2. Use Ampere’s circuital law: $\oint \vec B\cdot d\vec l = \mu_0 I_{\text{enc}}$

3. Choose an Amperian loop of radius $r$ inside the wire. By symmetry, magnetic field $B$ is constant on this circle and tangent to it. So, $B(2\pi r)=\mu_0 I_{\text{enc}}$

4. **Find enclosed current $I_{\text{enc}}$: ** Since current is uniformly distributed, $\text{Current density } J = \frac{I}{\pi R^2}$ Area enclosed within radius $r$ is $\pi r^2$ Therefore, $I_{\text{enc}} = J(\pi r^2)=\frac{I}{\pi R^2}\cdot \pi r^2 = I\frac{r^2}{R^2}$

5. Substitute into Ampere’s law: $B(2\pi r)=\mu_0 I\frac{r^2}{R^2}$ $B=\frac{\mu_0 I}{2\pi R^2}r$

6. Hence, $B \propto r$

7. Option check:

   • A: $B\propto r^2$ ❌
   • B: $B\propto r$ ✅
   • C: $B\propto \dfrac{1}{r^2}$ ❌
   • D: $B\propto \dfrac{1}{r}$ ❌

Therefore, the correct answer is B.