Question

A thick current-carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by 

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is D

To determine the variation of the magnetic field \(B(r)\) due to a thick current-carrying cable with radius \(R\) and current \(I\), we need to consider both the regions inside and outside the cable:

1. Inside the Cable (\(r \leq R\))

The magnetic field inside a current-carrying conductor can be calculated using Ampere’s Law, which states:

\(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}\)

For a circle of radius \(r\) inside the cable, the enclosed current \(I_{\text{enc}}\) is proportional to the area of the circle:

\(I_{\text{enc}} = I \frac{\pi r^2}{\pi R^2} = I \left(\frac{r^2}{R^2}\right)\)

Using Ampere’s law, we provide:

\(B(r) \cdot 2\pi r = \mu_0 I \left(\frac{r^2}{R^2}\right)\)

Simplifying this, the formula for magnetic field \(B(r)\) inside the cable becomes:

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

2. Outside the Cable (\(r > R\))

For the region outside the cable, the entire current \(I\) is enclosed by the Ampere's loop, and Ampere's law becomes:

\(B(r) \cdot 2\pi r = \mu_0 I\)

Simplifying this, the magnetic field \(B(r)\) outside the cable is given by:

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

Conclusion:

• For \(r \leq R\), \(B(r) \propto r\) (linear variation with distance).
• For \(r > R\), \(B(r) \propto \frac{1}{r}\) (inverse variation with distance).

The correct graph that represents this behavior is the one that shows a linear increase within the radius and inversely proportional decrease outside the radius. Upon examining the options given, the correct option is: