COMEDK · Physics · 22. Electromagnetic Induction
Around the central part of an air cored solenoid of length \(20 \mathrm{~cm}\) and area of cross section \(1.4 \times 10^{-3} \mathrm{~m}^2\) and 3000 turns, another coil of 250 turns is closely wound. A current \(2 \mathrm{~A}\) in the solenoid is reversed in \(0.2 \mathrm{~s}\), then the induced emf produced is
- A \(1.32 \times 10^{-1} \mathrm{~V}\)
- B \(1.16 \times 10^{-1} \mathrm{~V}\)
- C \(4 \times 10^{-1} \mathrm{~V}\)
- D \(8 \times 10^{-2} \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(1.32 \times 10^{-1} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The magnetic field \(B\) inside a long solenoid is given by \(B = \mu_{0} n I\), where \(n = \dfrac{N}{l}\) is the number of turns per unit length. Given: \(N_{1} = 3000\), \(l = 0.2 \text{ m}\), \(I_{initial} = 2 \text{ A}\), \(I_{final} = -2 \text{ A}\),…
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