Two neighbouring coils A and B have a mutual inductance of 20 mH. The current flowing through A is given by, `i = 3t^(2) - 4t + 6`. The induced eml at t=2s is

Options

A. 160 mV
B. 200 mV
C. 260 mV
D. 300 mV

Answer

To find the induced electromotive force (emf) in coil B due to the changing current in coil A, we can use the formula for mutual inductance. The induced emf (E) in coil B is given by:

\[ E = -M \frac{dI}{dt} \]

where:
- \( M \) is the mutual inductance,
- \( \frac{dI}{dt} \) is the rate of change of current in coil A.

### Step-by-Step Solution:

1. **Identify the Mutual Inductance**:
Given that the mutual inductance \( M = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \).

2. **Write the Current Equation**:
The current flowing through coil A is given by:
\[ I(t) = 3t^2 - 4t + 6 \]

3. **Differentiate the Current with Respect to Time**:
We need to find \( \frac{dI}{dt} \):
\[ \frac{dI}{dt} = \frac{d}{dt}(3t^2 - 4t + 6) \]
Using the power rule of differentiation:
\[ \frac{dI}{dt} = 6t - 4 \]

4. **Evaluate \( \frac{dI}{dt} \) at \( t = 2 \, \text{s} \)**:
Substitute \( t = 2 \):
\[ \frac{dI}{dt} = 6(2) - 4 = 12 - 4 = 8 \, \text{A/s} \]

5. **Calculate the Induced EMF**:
Now, substitute \( M \) and \( \frac{dI}{dt} \) into the induced emf formula:
\[ E = -M \frac{dI}{dt} = - (20 \times 10^{-3}) (8) \]
\[ E = -160 \times 10^{-3} \, \text{V} = -0.16 \, \text{V} = -160 \, \text{mV} \]

6. **Conclusion**:
The induced emf at \( t = 2 \, \text{s} \) is \( -160 \, \text{mV} \).