Question

A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre

• A. \(B∝r^2\)
• B. \(B∝r\)
• C. \(B∝\frac {1}{r^2}\)
• D. \(B∝\frac {1}{r}\)

Answer 1

The Correct Option is B

 To solve this problem, we need to determine how the magnetic field \(B\) varies with the distance \(r\) from the centre of a long straight wire with a uniform current distribution across its cross-section. The wire has a radius \(R\), and we are interested in the region where \(r < R\).

Concepts Used:

• According to Ampère's Law, the magnetic field \(\textbf{B}\) around a current carrying conductor can be found using the relation:
• \(\oint \mathbf{B} \cdot \,d\mathbf{l} = \mu_0 I_{\text{enclosed}}\)
• For a point inside the wire, \(I_{\text{enclosed}}\) is the current enclosed by a circle of radius \(r\), within the wire.

Step-by-Step Solution:

1. The current density \(J\) is defined as the current per unit area, given by:
2. \(J = \frac{I}{\pi R^2}\)
3. For a circle of radius \(r\), within the cross-section, the current enclosed \(I_{\text{enclosed}}\) is:
4. \(I_{\text{enclosed}} = J \times \pi r^2 = \frac{I}{\pi R^2} \times \pi r^2 = I \frac{r^2}{R^2}\)
5. Using Ampère's Law, for the circular path of radius \(r\):
6. \(\oint \mathbf{B} \cdot \,d\mathbf{l} = B \times 2\pi r = \mu_0 I_{\text{enclosed}}\)
7. Substitute \(I_{\text{enclosed}}\) from step 4, and solve for \(B\):
8. \(B \times 2\pi r = \mu_0 I \frac{r^2}{R^2}\)
9. Simplify to get the expression for \(B\):
10. \(B = \frac{\mu_0 I}{2\pi R^2} r\)

From the above expression, it is clear that the magnetic field \(B\) is directly proportional to \(r\) when \(r < R\).

Conclusion:

Therefore, the correct answer is that the magnetic field \(B\) varies linearly with distance \(r\) from the center of the wire within the radius \(R\), which confirms the correct option is:

• \(B∝r\) (Option: \(B∝r\))