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

Options

A. 2 Wb
B. 0.5 Wb
C. 12.5 Wb
D. Zero

Explanation

To find the magnetic flux linked with the coil, we can use the relationship between inductance (L), current (I), and magnetic flux (Φ). The formula that relates these quantities is:

\[ L = \frac{\Phi}{I} \]

Where:
- \( L \) is the inductance in Henrys (H),
- \( \Phi \) is the magnetic flux in Webers (Wb),
- \( I \) is the current in Amperes (A).

Given:
- Inductance \( L = 5 \, \text{H} \)
- Current \( I = 2.5 \, \text{A} \)

We can rearrange the formula to solve for magnetic flux \( \Phi \):

\[ \Phi = L \times I \]

Now, substituting the given values into the equation:

\[ \Phi = 5 \, \text{H} \times 2.5 \, \text{A} \]

Calculating the product:

\[ \Phi = 12.5 \, \text{Wb} \]

Thus, the magnetic flux linked with the coil is \( 12.5 \, \text{Wb} \).

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