Question

A 30 cm long solenoid has 10 turns per cm and area of 5 cm². The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is \(\alpha \times 10^{-5}\) V. The value \(\alpha\) is ________.

• A. 60
• B. 12
• C. 120
• D. 34

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A change in current through a solenoid induces an electromotive force (e.m.f.) due to self-induction. The induced e.m.f. is proportional to the rate of change of magnetic flux, which in turn is proportional to the rate of change of current.
Step 2: Key Formula or Approach:
1. Self-inductance $L = \mu_0 n^2 A l$.
2. Induced e.m.f. $|\varepsilon| = L \frac{di}{dt}$.
3. $n$ is turns per unit length, $l$ is length, $A$ is area.
Step 3: Detailed Explanation:
Given: $l = 0.3 \text{ m}$, $n = 10 \text{ turns/cm} = 1000 \text{ turns/m}$, $A = 5 \times 10^{-4} \text{ m}^2$. Change in current $\Delta I = 4 - 2 = 2 \text{ A}$, time $\Delta t = 3.14 \text{ s} \approx \pi \text{ s}$. \[ L = \mu_0 n^2 A l = (4\pi \times 10^{-7}) \times (1000)^2 \times (5 \times 10^{-4}) \times 0.3 \] \[ L = 4\pi \times 10^{-7} \times 10^6 \times 1.5 \times 10^{-4} = 6\pi \times 10^{-5} \text{ H} \] Now find e.m.f.: \[ |\varepsilon| = L \frac{\Delta I}{\Delta t} = (6\pi \times 10^{-5}) \times \frac{2}{\pi} \] The $\pi$ cancels out: \[ |\varepsilon| = 12 \times 10^{-5} \text{ V} \] Comparing with $\alpha \times 10^{-5} \text{ V}$, we find $\alpha = 12$.
Step 4: Final Answer:
The value of \(\alpha\) is 12.