In a coil, the current changes from $-2 \mathrm{~A}$ to $+2 \mathrm{~A}$ in $0.2 \mathrm{~s}$ and induces an emf of $0.1 \mathrm{~V}$. The self inductance of the coil is :

<p>To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:</p> <p>$\text{emf} = - L \frac{\Delta I}{\Delta t}$</p> <p>Where:</p> <ul> <li>$ \text{emf} $ = induced voltage in volts (V)</li> <li>$ L $ = self-inductance of the coil in henries (H)</li> <li>$ \Delta I $ = change in current in amperes (A)</li> <li>$ \Delta t $ = time interval in seconds (s) over which the current change occurs</li> </ul> <p>Here, the problem gives us the following data:</p> <ul> <li>$ \text{emf} = 0.1 \, \text{V} $</li> <li>$ \Delta I = 2 \, \text{A} - (-2 \, \text{A}) = 4 \, \text{A} $</li> <li>$ \Delta t = 0.2 \, \text{s} $</li> </ul> <p>Substituting the given values into the formula, we get:</p> <p>$0.1 = -L \frac{4}{0.2}$</p> <p>Solving for $ L $, the equation becomes:</p> <p>$0.1 = -L \cdot 20$</p> <p>Therefore:</p> <p>$L = - \frac{0.1}{20}$</p> <p>Calculating the value of $ L $:</p> <p>$L = -0.005 \, \text{H}$</p> <p>Or, expressing $ L $ in millihenries (mH):</p> <p>$L = -5 \, \text{mH}$</p> <p>However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):</p> <p>$L = 5 \, \text{mH}$</p> <p>Thus, the self-inductance of the coil is <strong>5 mH</strong>, which corresponds to <strong>Option D</strong>.</p>