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 emf at t = 2s is

Options

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

Explanation

To solve the problem, we need to find the induced emf in coil B due to the changing current in coil A. The formula for the induced emf (E) in coil B due to the current in coil A 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 1: Identify the given values
- Mutual inductance \( M = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- Current \( i(t) = 3t^2 - 4t + 6 \)

### Step 2: Differentiate the current with respect to time
To find \( \frac{dI}{dt} \), we differentiate the current function \( i(t) \):

\[ \frac{dI}{dt} = \frac{d}{dt}(3t^2 - 4t + 6) \]

Using the power rule of differentiation:

\[ \frac{dI}{dt} = 6t - 4 \]

### Step 3: Evaluate \( \frac{dI}{dt} \) at \( t = 2 \, \text{s} \)
Now, substitute \( t = 2 \) into the derivative:

\[ \frac{dI}{dt} \bigg|_{t=2} = 6(2) - 4 = 12 - 4 = 8 \, \text{A/s} \]

### Step 4: Calculate the induced emf
Now we can substitute \( M \) and \( \frac{dI}{dt} \) into the induced emf formula:

\[ E = -M \frac{dI}{dt} = - (20 \times 10^{-3}) (8) \]

Calculating this gives:

\[ E = -160 \times 10^{-3} \, \text{V} = -0.160 \, \text{V} \]

Since we are interested in the magnitude of the induced emf:

\[ |E| = 160 \, \text{mV} \]

### Final Answer
The induced emf at \( t = 2 \, \text{s} \) is \( 160 \, \text{mV} \).

---