Question

When current in a coil changes from 5A to 2A in 0.1s then average voltage of 50V is produced. The self-inductance of the coil is

• A. 1.67 henry
• B. 6 henry
• C. 3 henry
• D. 0.67 henry

Hint:

In problems involving induced e.m.f., always pay attention to the signs. The negative sign in Faraday's law (\(\epsilon = -L \frac{dI}{dt}\)) represents Lenz's law, which states that the induced e.m.f. opposes the change in current. When calculating the magnitude of inductance, you can often ignore the sign, but understanding its meaning is crucial for conceptual questions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question asks for the self-inductance of a coil. Self-inductance is the property of a coil by virtue of which it opposes any change in the strength of the current flowing through it by inducing an electromotive force (e.m.f.) in itself. This induced e.m.f. is also known as back e.m.f. The magnitude of this induced e.m.f. is directly proportional to the rate of change of current.
Step 2: Key Formula or Approach:
The formula for the average induced voltage (e.m.f., \(\epsilon\)) in a coil due to self-inductance (L) is given by Faraday's law of induction:
\[ \epsilon = -L \frac{dI}{dt} \] Where:
\(\epsilon\) is the induced voltage (e.m.f.).
\(L\) is the self-inductance of the coil.
\(\frac{dI}{dt}\) is the rate of change of current.
For average values over a time interval \(\Delta t\), the formula can be written as:
\[ \epsilon_{avg} = -L \frac{\Delta I}{\Delta t} \] The negative sign indicates that the induced e.m.f. opposes the change in current (Lenz's Law). For calculating the magnitude, we can use:
\[ |\epsilon_{avg}| = L \left| \frac{\Delta I}{\Delta t} \right| \] Step 3: Detailed Explanation:
Given data:
Initial current, \(I_1 = 5 \text{ A}\)
Final current, \(I_2 = 2 \text{ A}\)
Time interval, \(\Delta t = 0.1 \text{ s}\)
Average induced voltage, \(\epsilon = 50 \text{ V}\)
First, we calculate the change in current (\(\Delta I\)):
\[ \Delta I = I_2 - I_1 = 2 \text{ A} - 5 \text{ A} = -3 \text{ A} \] Next, we calculate the rate of change of current \(\frac{\Delta I}{\Delta t}\):
\[ \frac{\Delta I}{\Delta t} = \frac{-3 \text{ A}}{0.1 \text{ s}} = -30 \text{ A/s} \] Now, we use the formula for induced voltage to find the self-inductance \(L\):
\[ \epsilon = -L \frac{\Delta I}{\Delta t} \] \[ 50 \text{ V} = -L (-30 \text{ A/s}) \] \[ 50 = 30L \] Solving for \(L\):
\[ L = \frac{50}{30} \text{ H} \] \[ L = \frac{5}{3} \text{ H} \] \[ L \approx 1.67 \text{ H} \] Step 4: Final Answer:
The self-inductance of the coil is approximately 1.67 Henry. This corresponds to option (A).