QJEEM24Ph968b: JEE Main 2024 Physics - Class 12 JEEM24Ph968b Solution: In a coil, the current changes from − 2   A to + 2  …

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:

$$\text{emf}=-L\frac{\mathrm{\Delta }I}{\mathrm{\Delta }t}$$

Where:

• $\text{emf}$ = induced voltage in volts (V)
• $L$ = self-inductance of the coil in henries (H)
• $\Delta I$ = change in current in amperes (A)
• $\Delta t$ = time interval in seconds (s) over which the current change occurs

Here, the problem gives us the following data:

• $\text{emf} = 0.1 \, \text{V}$
• $\Delta I = 2 \, \text{A} - (-2 \, \text{A}) = 4 \, \text{A}$
• $\Delta t = 0.2 \, \text{s}$

Substituting the given values into the formula, we get:

$$0.1=-L\frac{4}{0.2}$$

Solving for $L$ , the equation becomes:

$$0.1=-L\cdot 20$$

Therefore:

$$L=-\frac{0.1}{20}$$

Calculating the value of $L$ :

$$L=-0.005\text{H}$$

Or, expressing $L$ in millihenries (mH):

$$L=-5\text{mH}$$

However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):

$$L=5\text{mH}$$

Thus, the self-inductance of the coil is 5 mH, which corresponds to Option D.