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 C

To solve this problem, we need to determine how the magnetic field \(B(r)\) varies with distance \(r\) from the axis of a thick cable carrying a uniformly distributed current \(I\).

The situation involves using Ampère's Law, which states that the line integral of the magnetic field \(B\) around a closed path is equal to \(\mu_0\) times the current enclosed by that path:

According to Ampère’s Law:

\[\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}\]

where \(I_{\text{enc}}\) is the current enclosed by the loop.

1. If \(r < R\) (inside the cable):
   • The current density \(J\) is uniform, thus \(J = \frac{I}{\pi R^2}\).
   • The current enclosed, \(I_{\text{enc}} = J \cdot \pi r^2 = \frac{I r^2}{R^2}\).
   • Apply Ampère’s Law: \(B \cdot 2\pi r = \mu_0 \frac{I r^2}{R^2}\).
   • Thus, \(B = \frac{\mu_0 I r}{2\pi R^2}\).
2. If \(r \ge R\) (outside the cable):
   • All the current \(I\) is enclosed by the Amperian loop.
   • Apply Ampère’s Law: \(B \cdot 2\pi r = \mu_0 I\).
   • Thus, \(B = \frac{\mu_0 I}{2\pi r}\).

The correct plot of \(B(r)\) vs \(r\) should show:

• For \(r < R\), \(B \propto r\) (linearly increasing).
• For \(r \ge R\), \(B \propto \frac{1}{r}\) (hyperbolically decreasing).

Thus, the correct choice is the plot which matches these characteristics. From the options provided, the correct visual representation is:

This plot accurately demonstrates the linear increase of \(B(r)\) inside the cable and the hyperbolic decrease outside it.