Question

A long solenoid having \(100\) turns per cm carries a current of \(\dfrac{4}{\pi}\,\text{A}\). At the centre of it is placed a coil of \(200\) turns of cross-sectional area \(25\,\text{cm}^2\) having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed within \(0.04\,\text{s}\), the induced emf in the coil is:

• A. \(0.2\,\text{V}\)
• B. \(0.4\,\text{V}\)
• C. \(0.002\,\text{V}\)
• D. \(0.016\,\text{V}\)

Hint:

When current in a solenoid is reversed, the magnetic field changes from \(+B\) to \(-B\). Therefore, the change in magnetic field is \(2B\), not just \(B\).

Answer 1

The Correct Option is B

Step 1: Find the magnetic field and initial flux linkage.
\( n = 100\,\text{turns/cm} = 10^4\,\text{turns/m} \). \( B = \mu_0 nI = 4\pi\times10^{-7}\times10^4\times\frac{4}{\pi} = 16\times10^{-3}\,\text{T} \). Flux linkage in inner coil: \( N\Phi = 200\times16\times10^{-3}\times25\times10^{-4} = 8\times10^{-3}\,\text{Wb} \).

Step 2: Find induced EMF when current reverses.
Change in flux linkage \( = 2\times8\times10^{-3}\,\text{Wb} \). \[ \varepsilon = \frac{\Delta(N\Phi)}{\Delta t} = \frac{16\times10^{-3}}{0.04} = 0.4\,\text{V} \] \[ \boxed{0.4\,\text{V}} \]