Solution: Inductance of Folded Copper Foil
1. Magnetic Field CalculationThe current flows uniformly into the edge A and out of edge B across the width \(b\). This configuration creates a sheet of current.
- For the flat portion, the two parallel plates carrying opposite currents behave like a large planar solenoid or transmission line. Applying Ampere’s Law: \(B b = \mu_0 I \implies B = \frac{\mu_0 I}{b}\).
- For the cylindrical portion, it behaves like a single-turn solenoid of length \(b\). The field inside is \(B = \mu_0 n I\). Since it is one turn over length \(b\), the effective turn density is \(1/b\) (or strictly, the linear current density is \(K = I/b\)). Thus, \(B = \frac{\mu_0 I}{b}\).
The total magnetic flux \(\Phi\) is the product of the magnetic field and the total cross-sectional area enclosed by the foil. The enclosed area consists of two parts: 1. The area of the cylindrical section: \(A_{cylinder} = \pi r^2\) 2. The area between the flat plates: \(A_{flat} = a \times d\) Total Area \(A = \pi r^2 + ad\). Therefore, the total flux is: $$ \Phi = B \cdot A = \left( \frac{\mu_0 I}{b} \right) (\pi r^2 + ad) $$
3. InductanceThe inductance \(L\) is defined as flux per unit current (\(L = \Phi / I\)):