Question

A solenoid having length 30 cm. If there are 10 turns/cm and current through solenoid changes from 2A to 4A in 3.14 sec. Find emf induced. (Area is A)

• A. \(0.24\,A\ \text{volt}\)
• B. \(0.40\,A\ \text{volt}\)
• C. \(0.80\,A\ \text{volt}\)
• D. \(0.20\,A\ \text{volt}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The induced emf in a solenoid is caused by the change in current passing through it.
This is governed by Faraday's law of induction.
Step 2: Key Formula or Approach:
Induced emf is given by $e = L \frac{\Delta I}{\Delta t}$.
Self-inductance of a solenoid is $L = \mu_0 n^2 A l$, where $n$ is turns per unit length.
Step 3: Detailed Explanation:
Given parameters:
Length $l = 30 \text{ cm} = 0.3 \text{ m}$.
Turns per unit length $n = 10 \text{ turns/cm} = 1000 \text{ turns/m}$.
Change in current $\Delta I = 4 \text{ A} - 2 \text{ A} = 2 \text{ A}$.
Time interval $\Delta t = 3.14 \text{ s} \approx \pi \text{ s}$.
First, calculate the self-inductance $L$:
\[ L = (4\pi \times 10^{-7}) \times (1000)^2 \times A \times 0.3 \]
\[ L = 4\pi \times 10^{-7} \times 10^6 \times 0.3 \times A = 1.2\pi \times 10^{-1} A \text{ Henry} \]
Now, calculate the induced emf $e$:
\[ e = L \frac{\Delta I}{\Delta t} = (1.2\pi \times 10^{-1} A) \times \frac{2}{3.14} \]
Since $\pi \approx 3.14$, they cancel each other out:
\[ e = 1.2 \times 10^{-1} A \times 2 = 0.24 A \text{ volts} \]
Step 4: Final Answer:
The induced emf is $0.24 A$ volt.