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A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is

1. 2 Wb
2. 0.5 Wb
3. 12.5 Wb
4. Zero

Correct answer: 12.5 Wb

Solution

The magnetic flux linked with a coil is given by the formula \( \Phi = L \cdot I \), where \( L \) is the inductance and \( I \) is the current. Substituting \( L = 5 \) H and \( I = 2.5 \) A, we get \( \Phi = 5 \cdot 2.5 = 12.5 \) Wb.

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