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 the problem of the magnetic field variation inside a thick current-carrying cable, where the current is uniformly distributed across the cross-section, we will use the principles of magnetism and apply Ampere's Law.

Understanding the Problem:

We have a current-carrying cable with radius $R$ and a uniformly distributed current $I$. We need to determine how the magnetic field $B(r)$ changes with the distance $r$ from the cable's axis.

Theory and Formulae:

By Ampère's Law, the line integral of the magnetic field $B$ around a closed path is equal to $\mu_0$ times the current passing through the enclosed area:

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

Here, $I_{enc}$ is the current enclosed by the path. For a radial distance $r \leq R$, the magnitude of the magnetic field $B(r)$ inside the cable can be derived as follows:

• $dI = I \cdot \left( \frac{\pi r^2}{\pi R^2} \right) = I \cdot \frac{r^2}{R^2}$, the current enclosed by a circle of radius $r$.
• Thus, $B(r) \cdot 2 \pi r = \mu_0 \cdot \frac{I r^2}{R^2}$
• Therefore, $ B(r) = \frac{\mu_0 I r}{2 \pi R^2} $

Conclusion:

The magnetic field inside the cable increases linearly with the distance $r$ from the axis of the cable, and it is directly proportional to $ \frac{\mu_0 I}{R^2} $.

Hence, the correct option for the variation of magnetic field $B(r)$ inside the cable versus the distance $r$ is as shown in the above diagram.