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:
When current passing through a solenoid changes, it induces an electromotive force (e.m.f.) due to self-induction. We first need to calculate the self-inductance $L$ of the solenoid, and then apply Faraday's law of induction.
Step 2: Key Formula or Approach:
1. Self-inductance of a solenoid: $L = \mu_0 n^2 A l$, where $n$ is turns per unit length, $A$ is cross-sectional area, and $l$ is length.
2. Induced e.m.f.: $|e| = L \frac{\Delta I}{\Delta t}$.
Step 3: Detailed Explanation:
Given values:
Length $l = 30\text{ cm} = 0.3\text{ m}$.
Turns per unit length $n = 10\text{ turns/cm} = 1000\text{ turns/m}$.
Cross-sectional area $A = 5\text{ cm}^2 = 5 \times 10^{-4}\text{ m}^2$.
Current change $\Delta I = 4\text{ A} - 2\text{ A} = 2\text{ A}$.
Time interval $\Delta t = 3.14\text{ s} \approx \pi\text{ s}$.
Permeability of free space $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$.
First, calculate the self-inductance $L$:
$L = (4\pi \times 10^{-7}) \times (1000)^2 \times (5 \times 10^{-4}) \times (0.3)$.
$L = 4\pi \times 10^{-7} \times 10^6 \times 1.5 \times 10^{-4}$.
$L = 4\pi \times 1.5 \times 10^{-5} = 6\pi \times 10^{-5}\text{ H}$.
Now, calculate the induced e.m.f.:
$|e| = L \frac{\Delta I}{\Delta t} = (6\pi \times 10^{-5}) \times \frac{2}{3.14}$.
Since $\pi \approx 3.14$, we can cancel them out:
$|e| = 6 \times 10^{-5} \times 2 = 12 \times 10^{-5}\text{ V}$.
The problem states the e.m.f. is $\alpha \times 10^{-5}\text{ V}$.
Comparing the expressions, we get $\alpha = 12$.
Step 4: Final Answer:
The value of $\alpha$ is 12.