Question

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

• A. \( 5 \, \text{mH} \)
• B. \( 1 \, \text{mH} \)
• C. \( 2.5 \, \text{mH} \)
• D. \( 4 \, \text{mH} \)

Hint:

To calculate self-inductance, ensure accurate substitution of the total current change and time interval. The relationship \( L = \frac{\mathcal{E} \cdot \Delta t}{\Delta I} \) directly links the rate of current change to the induced emf.

Answer 1

The Correct Option is A

The induced emf in a coil is described by the equation: \[ \mathcal{E} = L \frac{\Delta I}{\Delta t}, \] where: \( \mathcal{E} = 0.1 \, \text{V} \) is the induced emf, \( \Delta I = I_{\text{final}} - I_{\text{initial}} = 2 - (-2) = 4 \, \text{A} \) is the change in current, \( \Delta t = 0.2 \, \text{s} \) is the time interval, \( L \) is the self-inductance of the coil. Step 1: Rearrange to Solve for \( L \) Rearrange the formula for \( L \): \[ L = \frac{\mathcal{E} \cdot \Delta t}{\Delta I}. \] Substitute the given values: \[ L = \frac{0.1 \cdot 0.2}{4}. \] Simplify: \[ L = \frac{0.02}{4} = 0.005 \, \text{H}. \] Convert \( L \) to millihenries: \[ L = 5 \, \text{mH}. \] Final Answer: \[ \boxed{5 \, \text{mH}} \]