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