Question

As frequency increases, the surface resistance of a metal :

• A. Decreases
• B. Increases
• C. Remains unchanged
• D. Varies in an unpredictable manner

Hint:

Due to skin effect: \[ R_s\propto \sqrt{f} \] Higher frequency causes current crowding near conductor surface, increasing AC resistance.

Answer 1

The Correct Option is B

Concept: At high frequencies, alternating current does not penetrate uniformly throughout the conductor. Instead:
• Current tends to flow near the surface.
• This phenomenon is called skin effect. Because of skin effect:
• Effective conducting area decreases.
• Surface resistance increases with frequency. The surface resistance of a conductor is given by: \[ R_s=\sqrt{\frac{\omega \mu}{2\sigma}} \] where:
• \(\omega=2\pi f\) is angular frequency
• \(\mu\) is permeability
• \(\sigma\) is conductivity From the formula: \[ R_s \propto \sqrt{f} \] Thus surface resistance increases with frequency.

Step 1: Understand the skin effect. When frequency increases:
• Current crowding near conductor surface increases.
• Inner region of conductor carries less current. Hence: \[ \text{Effective area decreases} \]

Step 2: Relate effective area to resistance. Resistance is inversely proportional to conducting area: \[ R\propto \frac{1}{A} \] Since effective area decreases: \[ R_s \uparrow \] Therefore surface resistance increases.

Step 3: Use the mathematical formula. Surface resistance: \[ R_s=\sqrt{\frac{\omega \mu}{2\sigma}} \] Since: \[ \omega=2\pi f \] we obtain: \[ R_s\propto \sqrt{f} \] Thus: \[ \text{Higher frequency} \Rightarrow \text{Higher surface resistance} \]

Step 4: Write the final answer. Therefore, as frequency increases: \[ \boxed{\text{Surface resistance increases}} \] Hence, the correct option is: \[ \boxed{(B)\ \text{Increases}} \]