Question

When a current changes at the rate of $30\ \text{A s}^{-1}$, the induced emf is $12\ \text{V}$. The self-inductance of the coil is:

• A. 0.4 H
• B. 0.2 H
• C. 0.6 H
• D. 0.3 H
• E. 0.1 H

Hint:

$L = \text{emf} / (\text{rate of change of current})$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem relates the induced electromotive force (emf) in a coil to the rate of change of current flowing through it. This relationship is defined by the coil's self-inductance.
Step 2: Key Formula or Approach:
The self-induced emf (\( \mathcal{E} \)) in a coil is given by Faraday's law of induction for an inductor: \[ \mathcal{E} = -L \frac{dI}{dt} \] where L is the self-inductance and \( \frac{dI}{dt} \) is the rate of change of current. The negative sign (Lenz's Law) indicates that the induced emf opposes the change in current. For calculating the magnitude, we can use: \[ |\mathcal{E}| = L \left| \frac{dI}{dt} \right| \] We need to rearrange this to solve for L. \[ L = \frac{|\mathcal{E}|}{|dI/dt|} \] Step 3: Detailed Explanation:
We are given: - Magnitude of the induced emf, \( |\mathcal{E}| = 12 \text{ V} \) - Rate of change of current, \( \left| \frac{dI}{dt} \right| = 30 \text{ A/s} \) Substitute these values into the formula for L: \[ L = \frac{12}{30} \] Simplify the fraction: \[ L = \frac{12 \div 6}{30 \div 6} = \frac{2}{5} = 0.4 \text{ H} \] Step 4: Final Answer:
The self-inductance of the coil is 0.4 H.