Question

If the energy stored in an inductor is 18 mJ when a current of 3 A is passed through it, then the magnetic flux linked with the inductor is

• A. \(36\ \text{mWb}\)
• B. \(24\ \text{mWb}\)
• C. \(18\ \text{mWb}\)
• D. \(12\ \text{mWb}\) \bigskip

Hint:

Remember the relation \[ U=\frac12 I\lambda \] Thus, \[ \lambda=\frac{2U}{I} \] This formula provides a direct shortcut for such questions.

Answer 1

The Correct Option is D

Concept: An inductor stores electrical energy in the form of magnetic energy when current flows through it. The energy stored in an inductor is given by \[ U=\frac12 LI^2 \] where \[ L=\text{Inductance}, \qquad I=\text{Current} \] Also, magnetic flux linkage is related to inductance by \[ \lambda = LI \] Using these two relations, we can determine the magnetic flux linked with the inductor.

Step 1: Write the given data.
Energy stored: \[ U=18\,\text{mJ}=18\times10^{-3}\,\text{J} \] Current: \[ I=3\,\text{A} \]

Step 2: Calculate the inductance of the coil.
Using \[ U=\frac12 LI^2 \] Substituting the given values, \[ 18\times10^{-3} =\frac12 L(3)^2 \] \[ 18\times10^{-3} =\frac92L \] \[ L=\frac{36\times10^{-3}}{9} \] \[ L=4\times10^{-3}\,\text{H} \] \[ L=4\,\text{mH} \]

Step 3: Determine the magnetic flux linkage.
\[ \lambda =LI \] \[ =(4\times10^{-3})(3) \] \[ =12\times10^{-3}\,\text{Wb} \] \[ =12\,\text{mWb} \]

Step 4: Write the final answer.
Hence the magnetic flux linked with the inductor is \[ \boxed{12\,\text{mWb}} \]