Question

The current passing through a coil of 120 turns and inductance \( 40 \) mH is \( 30 \) mA. The magnetic flux linked with the coil is:

• A. \( 20 \times 10^{-6} \) Wb
• B. \( 5 \times 10^{-6} \) Wb
• C. \( 12 \times 10^{-6} \) Wb
• D. \( 10 \times 10^{-6} \) Wb

Hint:

Magnetic flux \( \phi \) is given by \( \phi = L I \), where \( L \) is inductance and \( I \) is current.

Answer 1

The Correct Option is D

Step 1: Formula for Magnetic Flux The magnetic flux \( \Phi \) linked with the coil is given by: \[ \Phi = L \cdot I \] where:
- \( L \) is the inductance of the coil (given as \( 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} \)),
- \( I \) is the current passing through the coil (given as \( 30 \, \text{mA} = 30 \times 10^{-3} \, \text{A} \)). Step 2: Substituting Values Substitute the given values into the formula for magnetic flux: \[ \Phi = (40 \times 10^{-3}) \times (30 \times 10^{-3}) = 1200 \times 10^{-6} \, \text{Wb} \] \[ \Phi = 10 \times 10^{-6} \, \text{Wb} \] Step 3: Conclusion Thus, the magnetic flux linked with the coil is \( \boxed{10 \times 10^{-6} \, \text{Wb}} \).

Answer 2

Given:
- Number of turns, \( N = 120 \)
- Inductance, \( L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} \)
- Current, \( I = 30 \, \text{mA} = 30 \times 10^{-3} \, \text{A} \)

We need to find the magnetic flux \( \Phi \) linked with the coil.

Step 1: Relation between inductance and flux linkage:
The inductance \( L \) of a coil is related to the magnetic flux linkage \( \lambda \) and current \( I \) by:
\[ L = \frac{\lambda}{I} \] where the flux linkage \( \lambda = N \Phi \), and \( \Phi \) is the magnetic flux through one turn.

Step 2: Rearranging for flux \( \Phi \):
\[ L = \frac{N \Phi}{I} \Rightarrow \Phi = \frac{L I}{N} \]

Step 3: Substitute the given values:
\[ \Phi = \frac{40 \times 10^{-3} \times 30 \times 10^{-3}}{120} = \frac{1.2 \times 10^{-3}}{120} = 10 \times 10^{-6} \, \text{Wb} \]

Therefore, the magnetic flux linked with the coil is:
\[ \boxed{10 \times 10^{-6} \, \text{Wb}} \]