A solenoid of 2000 turns is wound over a length of 0.3 m. The area aof cross section is `1.3 xx 10^(-3) m^(2)`. Around its central seciton, a coil of 300 turns is closely wound. If an initial current of 2 A is reversed in 0.25 s, find the e.m.f induced in the coil .

Options

A. `6 xx 10^(-4)V`
B. `4.8 xx 10^(-2) V`
C. ` 2.4 xx 10^(-2) v`
D. 48kV

Answer

To find the e.m.f. induced in the coil wound around a solenoid, we can use the formula for mutual inductance. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Number of turns in the solenoid (N1) = 2000 turns
- Length of the solenoid (L) = 0.3 m
- Cross-sectional area of the solenoid (A) = \(1.3 \times 10^{-3} \, m^2\)
- Number of turns in the coil (N2) = 300 turns
- Initial current (I_initial) = 2 A
- Final current (I_final) = -2 A (since the current is reversed)
- Time taken for the reversal (Δt) = 0.25 s

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

### Step 3: Calculate di/dt
Now, we can find the rate of change of current (di/dt):
\[ \frac{di}{dt} = \frac{di}{\Delta t} = \frac{-4 \, A}{0.25 \, s} = -16 \, A/s \]

### Step 4: Calculate the Mutual Inductance (M)
The mutual inductance (M) between the solenoid and the coil can be calculated using the formula:
\[ M = \frac{\mu_0 N_1 N_2 A}{L} \]
Where:
- \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A\) (permeability of free space)
- \(N_1 = 2000\)
- \(N_2 = 300\)
- \(A = 1.3 \times 10^{-3} \, m^2\)
- \(L = 0.3 \, m\)

Substituting the values:
\[ M = \frac{(4\pi \times 10^{-7}) \times 2000 \times 300 \times (1.3 \times 10^{-3})}{0.3} \]

Calculating the above expression:
\[ M = \frac{(4\pi \times 10^{-7}) \times 2000 \times 300 \times 1.3 \times 10^{-3}}{0.3} \]
\[ M \approx 4.8 \times 10^{-3} \, H \]

### Step 5: Calculate the Induced e.m.f. (ε)
The induced e.m.f. (ε) in the coil can be calculated using the formula:
\[ \epsilon = -M \frac{di}{dt} \]
Substituting the values:
\[ \epsilon = - (4.8 \times 10^{-3}) \times (-16) \]
\[ \epsilon = 4.8 \times 10^{-2} \, V \]

### Final Answer
The e.m.f. induced in the coil is:
\[ \epsilon = 4.8 \times 10^{-2} \, V \]

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