Design a 2 cm long “air cored” rod type inductor to have8 µF to within 5%.(b) (1 mark) Calculate the magnetic flux Φ in Wb through the coil from0.75 A DC
Question
Design a 2 cm long “air cored” rod type inductor to have8 µF to within 5%.(b) (1 mark) Calculate the magnetic flux Φ in Wb through the coil from0.75 A DC
Solution
The question seems to be a bit confused. An inductor (coil) is characterized by its inductance (L), not capacitance (C). The unit of inductance is the Henry (H), not the Farad (F) which is the unit of capacitance. Therefore, it's not possible to design an inductor to have 8 µF (microfarads) because µF is a unit of capacitance, not inductance.
However, if you meant to design an inductor with an inductance of 8 µH (microhenries), then the design would depend on the number of turns in the coil, the area of the coil, and the length of the coil. The inductance (L) of a coil can be calculated using the formula:
L = (μ0 * N^2 * A) / l
where:
- L is the inductance,
- μ0 is the permeability of free space (4π × 10^-7 H/m),
- N is the number of turns,
- A is the cross-sectional area (in m^2), and
- l is the length of the coil (in m).
As for the magnetic flux (Φ), it can be calculated using the formula:
Φ = B * A
where:
- Φ is the magnetic flux (in Wb),
- B is the magnetic field (in T), and
- A is the area (in m^2).
The magnetic field (B) in a coil can be calculated using the formula:
B = μ0 * (N / l) * I
where:
- B is the magnetic field,
- μ0 is the permeability of free space,
- N is the number of turns,
- l is the length of the coil, and
- I is the current (in A).
So, you would first calculate B using the given current (0.75 A), then use this B to calculate the magnetic flux (Φ).
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