Question

In a coil, the current changes form –2 A to +2A in 0.2 s and induces an emf of 0.1 V. The self-inductance of the coil is :

• A. 5 mH
• B. 1 mH
• C. 2.5 mH
• D. 4 mH

Answer 1

The Correct Option is A

To determine the self-inductance of the coil, we can use the following formula that relates the induced electromotive force (emf) in a coil to its self-inductance and the rate of change of current through it:

\(E = -L \frac{\Delta I}{\Delta t}\)

Where:

• \(E\) is the induced emf (0.1 V in this case).
• \(L\) is the self-inductance of the coil.
• \(\Delta I\) is the change in current (-2 A to +2 A, so \(\Delta I = 2 - (-2) = 4 \, \text{A}\)).
• \(\Delta t\) is the time interval over which the current changes (0.2 s).

Rearranging the formula to solve for \(L\), we have:

\(L = -\frac{E \cdot \Delta t}{\Delta I}\)

Substitute the given values into this formula:

\(L = -\frac{0.1 \, \text{V} \cdot 0.2 \, \text{s}}{4 \, \text{A}}\)

\(L = -\frac{0.02}{4}\)

\(L = -0.005 \, \text{H}\)

Since inductance is conventionally given as a positive value, we take the absolute value:

\(L = 0.005 \, \text{H} = 5 \, \text{mH}\)

Thus, the self-inductance of the coil is 5 mH.

Let's analyze the options:

• 5 mH: Correct, matches calculated value.
• 1 mH: Incorrect.
• 2.5 mH: Incorrect.
• 4 mH: Incorrect.

Therefore, the correct answer is 5 mH.

Answer 2

The induced emf in a coil is given by:\[(\text{Emf})_{\text{induced}} = -L \frac{di}{dt}\]
In terms of magnitude:
\[|\text{Emf}_{\text{induced}}| = \left| L \frac{di}{dt} \right|\]
Given:
\[|\text{Emf}_{\text{induced}}| = 0.1 \, \text{V}\]
\[\frac{di}{dt} = \frac{2 - (-2)}{0.2} = \frac{4}{0.2} = 20 \, \text{A/s}\]
Now, solve for \( L \):
\[0.1 = L \times 20\]
\[L = \frac{0.1}{20} = 0.005 \, \text{H} = 5 \, \text{mH}\]