A long solenoid having `200` turns per centimeter carries a current of `1.5 A`. At the center of the solenoid is placed a coil of `100` turns of cross-sectional area `3.14 xx 10^(-4) m^(2)` having its axis parallel to the field produced by the solenoid. When the direction of current in teh solenoid is reversed within `0.05 s`, the induced emf in the coil is

Options

A. `0.48 V`
B. `0.048 V`
C. `0.0048 V`
D. `48 V`

Explanation

To solve the problem step by step, we will follow the concepts of electromagnetism, particularly focusing on the solenoid's magnetic field, magnetic flux, and induced electromotive force (emf).

### Step 1: Calculate the Magnetic Field (B) inside the Solenoid
The formula for the magnetic field inside a long solenoid is given by:
\[ B = \mu_0 n I \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( n \) is the number of turns per unit length (in turns/meter),
- \( I \) is the current in amperes.

Given:
- \( n = 200 \, \text{turns/cm} = 20000 \, \text{turns/m} \) (since 1 cm = 0.01 m),
- \( I = 1.5 \, \text{A} \).

Substituting the values:
\[ B = (4\pi \times 10^{-7}) \times (20000) \times (1.5) \]
\[ B = 3.77 \times 10^{-3} \, \text{T} \]

### Step 2: Calculate the Magnetic Flux (Φ) through the Coil
The magnetic flux through the coil is given by:
\[ \Phi = B \cdot A \]
where:
- \( A \) is the cross-sectional area of the coil.

Given:
- \( A = 3.14 \times 10^{-4} \, \text{m}^2 \).

Substituting the values:
\[ \Phi = (3.77 \times 10^{-3}) \times (3.14 \times 10^{-4}) \]
\[ \Phi = 1.18 \times 10^{-6} \, \text{Wb} \]

### Step 3: Calculate the Change in Magnetic Flux (ΔΦ)
When the current in the solenoid is reversed, the magnetic field also reverses, leading to a change in magnetic flux. The change in flux is:
\[ \Delta \Phi = 2 \Phi \]
because the flux changes from \( \Phi \) to \( -\Phi \).

So:
\[ \Delta \Phi = 2 \times (1.18 \times 10^{-6}) = 2.36 \times 10^{-6} \, \text{Wb} \]

### Step 4: Calculate the Induced EMF (E) in the Coil
The induced emf is calculated using Faraday's law of electromagnetic induction:
\[ E = -N \frac{d\Phi}{dt} \]
where:
- \( N \) is the number of turns in the coil,
- \( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux.

Given:
- \( N = 100 \, \text{turns} \),
- \( dt = 0.05 \, \text{s} \).

Substituting the values:
\[ E = -100 \times \frac{2.36 \times 10^{-6}}{0.05} \]
\[ E = -100 \times 4.72 \times 10^{-5} \]
\[ E = -0.00472 \, \text{V} \]

### Final Result
The induced emf in the coil when the current in the solenoid is reversed is approximately:
\[ E \approx 0.048 \, \text{V} \]