Question:

A long solenoid has 1500 turns. When a current of 3.5 A flows through it, the magnetic flux linked with each turn of solenoid is $2.8 \times 10^{-3}$ weber. The self-inductance of solenoid is

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The total flux linkage $\Phi = LI$ always refers to the total flux through all $N$ turns.
Updated On: Apr 28, 2026
  • 1.2 H
  • 2.4 H
  • 3.6 H
  • 6 H
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The Correct Option is A

Solution and Explanation


Step 1: Identify the given values. The number of turns $N = 1500$, the current $I = 3.5$ A, and the magnetic flux linked with each turn $\phi = 2.8 \times 10^{-3}$ Wb.
Step 2: Calculate the total magnetic flux ($\Phi_{net}$) linked with the entire solenoid by multiplying the number of turns by the flux per turn. \[ \Phi_{net} = N \phi \]
Step 3: Substitute the given values into the formula. \[ \Phi_{net} = 1500 \times 2.8 \times 10^{-3} = 4.2 \text{ Wb} \]
Step 4: Use the relationship between total magnetic flux and self-inductance $L$, which is $\Phi_{net} = LI$. \[ L = \frac{\Phi_{net{I} \]
Step 5: Calculate the self-inductance by substituting the values. \[ L = \frac{4.2}{3.5} = 1.2 \text{ H} \]
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