Question

The self-inductance of a choke coil is 5 henry. The current through it is increasing at a rate of 2 AS\(^{-1}\). The self-induced emf in the choke coil will be

• A. 2.5 V
• B. 5 V
• C. - 10 V
• D. 10 V

Hint:

Remember Lenz's law is represented by the negative sign in the formula \(\epsilon = -L \frac{dI}{dt}\). If the question asks for the \textit{magnitude} of the induced emf, the answer would be 10 V. However, since -10 V is an option, it is the more precise answer as it includes the direction (polarity) of the emf.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to Faraday's law of induction and Lenz's law, when the current through an inductor changes, a back electromotive force (emf) is induced across it. This self-induced emf opposes the change in current.
Step 2: Key Formula or Approach:
The self-induced emf (\(\epsilon\)) in an inductor is given by the formula: \[ \epsilon = -L \frac{dI}{dt} \] where:
\(L\) is the self-inductance of the coil.
\(\frac{dI}{dt}\) is the rate of change of current.
The negative sign indicates that the induced emf opposes the change in current (Lenz's Law).
Step 3: Detailed Explanation:
Given values are:
Self-inductance, \(L = 5\) H.
Rate of increase of current, \(\frac{dI}{dt} = 2\) A/s.
Substitute these values into the formula: \[ \epsilon = - (5 \, \text{H}) \times (2 \, \text{A/s}) \] \[ \epsilon = -10 \, \text{V} \] Step 4: Final Answer:
The self-induced emf in the choke coil is -10 V. The negative sign signifies that the emf is induced in a direction that opposes the increase in current. Therefore, option (C) is the correct answer.