A solenoid has 2000 turns would over a length of 0.30 m. The area of its cross-section is `1.2xx10^(-3)m^(2)`. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is

Options

A. `6xx10^(-4)V`
B. `4.8xx10^(-3)V`
C. `6xx10^(-2)V`
D. `32.1xx10^(-2)V`

Explanation

To solve the problem of finding the induced EMF in a solenoid when the current is reversed, we can follow these steps:

### Step 1: Identify the given values
- Number of turns (N) = 2000
- Length of the solenoid (l) = 0.30 m
- Cross-sectional area (A) = \(1.2 \times 10^{-3} \, m^2\)
- Initial current (I_initial) = 2 A
- Final current (I_final) = -2 A (since the current is reversed)
- Time interval (t) = 0.25 s

### Step 2: Calculate the change in current (di)
The change in current (di) can be calculated as:
\[ di = I_{final} - I_{initial} = -2 \, A - 2 \, A = -4 \, A \]

### Step 3: Calculate the self-inductance (L) of the solenoid
The formula for the self-inductance (L) of a solenoid is given by:
\[ L = \mu_0 \frac{N^2 A}{l} \]
Where:
- \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A\) (permeability of free space)

Substituting the values:
\[ L = (4\pi \times 10^{-7}) \frac{(2000)^2 (1.2 \times 10^{-3})}{0.30} \]

Calculating \(N^2\):
\[ N^2 = 2000^2 = 4,000,000 \]

Now substituting into the formula:
\[ L = (4\pi \times 10^{-7}) \frac{4,000,000 \times 1.2 \times 10^{-3}}{0.30} \]

Calculating the numerator:
\[ 4,000,000 \times 1.2 \times 10^{-3} = 4800 \]

Now substituting back:
\[ L = (4\pi \times 10^{-7}) \frac{4800}{0.30} \]
\[ L = (4\pi \times 10^{-7}) \times 16000 \]
\[ L = 64\pi \times 10^{-4} \, H \approx 0.201 \, H \]

### Step 4: Calculate the induced EMF (ε)
The induced EMF can be calculated using the formula:
\[ \epsilon = -L \frac{di}{dt} \]

Substituting the values:
\[ \epsilon = -0.201 \, H \times \frac{-4 \, A}{0.25 \, s} \]
\[ \epsilon = 0.201 \times 16 = 3.216 \, V \]

### Final Answer
The induced EMF in the solenoid is approximately \(3.216 \, V\).

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