Question

If the current in a coil changes from \(2A\) to \(4A\) in \(0.1\,s\), inducing an EMF of \(20V\), find the self-inductance.

• A. \(0.5\,H\)
• B. \(1\,H\)
• C. \(2\,H\)
• D. \(4\,H\)

Hint:

Self-induced EMF comes from the changing magnetic flux a coil links with itself, given by $E = \frac{d(N\Phi)}{dt}$ where $N\Phi = Li$. Find the rate of change of current first, then use it to solve for $L$.

Answer 1

The Correct Option is B

Concept: The induced EMF in an inductor is given by the relation \[ E = L\frac{di}{dt} \] where \(E\) = induced EMF, \(L\) = self-inductance, \(\frac{di}{dt}\) = rate of change of current.

Step 1: Write the given values. Initial current \(i_1 = 2A\) Final current \(i_2 = 4A\) Time interval \[ dt = 0.1\,s \] Induced EMF \[ E = 20V \]

Step 2: Find the rate of change of current. \[ \frac{di}{dt} = \frac{4 - 2}{0.1} \] \[ \frac{di}{dt} = \frac{2}{0.1} = 20 \]

Step 3: Substitute into the EMF formula. \[ E = L\frac{di}{dt} \] \[ 20 = L(20) \] \[ L = 1\,H \]

Answer 2

Concept:

• Self-inductance is defined through the flux linked with a coil due to its own current: $N\Phi = Li$, so any change in current directly changes the flux linkage by $\Delta(N\Phi) = L\,\Delta i$.
• By the law of electromagnetic induction, the induced EMF equals the rate of change of this flux linkage, which gives the same working formula but starting from the flux definition instead of quoting it directly.

Step 1: Write the law of electromagnetic induction for the coil.
$E = \frac{\Delta(N\Phi)}{\Delta t}$

Step 2: Replace the flux linkage using the self-inductance definition $N\Phi = Li$.
$E = \frac{L\,\Delta i}{\Delta t}$

Step 3: Substitute the given values.
$\Delta i = 4 - 2 = 2\,A$, $\ \Delta t = 0.1\,s$, $\ E = 20\,V$
$20 = \frac{L(2)}{0.1}$

Step 4: Solve for $L$.
$20 = 20L$

Final Answer: $L = 1\,H$