Question

A current of $2 \text{ A}$ produces a magnetic flux of $10^{-3}$ weber per turn in a coil of $1000$ turns. Then, the self-inductance of the coil is

• A. $0.10 \text{ H}$
• B. $0.25 \text{ H}$
• C. $0.20 \text{ H}$
• D. $0.30 \text{ H}$
• E. $0.50 \text{ H}$

Hint:

Always multiply the flux given "per turn" by the total number of turns to get the total linkage before calculating inductance.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Self-inductance (\(L\)) is a property of a coil that dictates how much total magnetic flux is generated for a given amount of current.
Step 2: Key Formula or Approach:
The total flux linkages \(N\Phi\) are proportional to the current \(I\):
\[ N\Phi = LI \] Where \(N\) is the number of turns, \(\Phi\) is the flux per turn, \(L\) is the self-inductance, and \(I\) is the current.
Step 3: Detailed Explanation:
Given values:
Current \(I = 2 \text{ A}\)
Flux per turn \(\Phi = 10^{-3} \text{ Wb}\)
Number of turns \(N = 1000\)
Substitute these into the formula:
\[ L \times 2 = 1000 \times 10^{-3} \] \[ 2L = 10^3 \times 10^{-3} \] \[ 2L = 1 \] \[ L = 0.5 \text{ H} \] Step 4: Final Answer:
The self-inductance is 0.50 H.