Question

The current in a coil decreases from 5 A to zero in a time of 0.1 s. If an average emf of 200 V is induced, then the self inductance of the coil is

• A. 20 H
• B. 4 H
• C. 2 H
• D. 40 H

Hint:

Remember the formula for self-induced EMF: $\mathcal{E} = -L \frac{dI}{dt}$. When calculating the magnitude of $L$, use the absolute values of EMF and the rate of change of current. Ensure all units are in their standard SI forms (Volts, Amperes, Seconds, Henries).

Answer 1

The Correct Option is B

Step 1: Identify the Formula for Induced EMF in Terms of Self-Inductance
When the current in a coil changes, an electromotive force (emf) is induced in the coil itself. This phenomenon is called self-induction. The induced emf ($\mathcal{E}$) is directly proportional to the rate of change of current ($\frac{dI}{dt}$). The constant of proportionality is the self-inductance ($L$) of the coil. The formula for the induced emf is: \[ \mathcal{E} = -L \frac{dI}{dt} \] The negative sign indicates that the induced emf opposes the change in current (Lenz's Law). When calculating the magnitude of the self-inductance, we often use the magnitude of the emf and the rate of change of current: \[ |\mathcal{E}| = L \left| \frac{dI}{dt} \right| \] Step 2: Extract Given Values from the Problem
Given: \begin{itemize} \item Initial current, $I_{\text{initial}} = 5 \text{ A}$ \item Final current, $I_{\text{final}} = 0 \text{ A}$ \item Change in time, $dt = 0.1 \text{ s}$ \item Average induced emf, $|\mathcal{E}| = 200 \text{ V}$ \end{itemize} Step 3: Calculate the Rate of Change of Current (\( \frac{dI}{dt} \))
The change in current, $dI = I_{\text{final}} - I_{\text{initial}} = 0 \text{ A} - 5 \text{ A} = -5 \text{ A}$. The rate of change of current is: \[ \frac{dI}{dt} = \frac{-5 \text{ A}}{0.1 \text{ s}} = -50 \text{ A/s} \] The magnitude of the rate of change of current is $\left| \frac{dI}{dt} \right| = 50 \text{ A/s}$. Step 4: Compute the Self-Inductance (\( L \))
Using the magnitude form of the induced emf formula: \[ |\mathcal{E}| = L \left| \frac{dI}{dt} \right| \] Rearrange the formula to solve for $L$: \[ L = \frac{|\mathcal{E}|}{\left| \frac{dI}{dt} \right|} \] Substitute the values: \[ L = \frac{200 \text{ V}}{50 \text{ A/s}} \] \[ L = 4 \text{ H} \] Step 5: Analyze Options
\begin{itemize} \item Option (1): 20 H. Incorrect. \item Option (2): 4 H. Correct, as it matches our calculated self-inductance. \item Option (3): 2 H. Incorrect. \item Option (4): 40 H. Incorrect. \end{itemize}