Question
Download Solution PDFWhat will be self inductance of a coil of 200 turns in which a current of 2A produces a magnetic flux of 4 mWb?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
- Self-inductance: The property of the current-carrying coil that opposes or resists the change of current flowing through itself is called self-inductance.
It is calculated by:
\(L =N {ϕ \over i}\)
where N is the number of turns, ϕ is the magnetic flux, i is current, and L is the self-inductance.
CALCULATION:
Given that N = 200 turns; i = 2A; ϕ = 4 mWb = 4 × 10-3 Wb;
Magnetic flux \(L =N {ϕ \over i}\)
\(L = 200 × {4×10^{-3} \over2}\)
L = 0.4 H
So the correct answer is option 4.
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